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#### Question 1:

The equation of the directrix of a hyperbola is xy + 3 = 0. Its focus is (−1, 1) and eccentricity 3. Find the equation of the hyperbola.

Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$⇒\sqrt{{\left(x-\left(-1\right)\right)}^{2}+{\left(y-1\right)}^{2}}=3×\left(\frac{x-y+3}{\sqrt{2}}\right)$
Squaring both the sides, we get:
${\left(x+1\right)}^{2}+{\left(y-1\right)}^{2}=\frac{9}{2}{\left(x-y+3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1+{y}^{2}-2y+1=\frac{9}{2}\left({x}^{2}+{y}^{2}+9-2xy-6y+6x\right)\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x+2+2{y}^{2}-4y+2=9{x}^{2}+9{y}^{2}+81-18xy-54y+54x\phantom{\rule{0ex}{0ex}}⇒7{x}^{2}+7{y}^{2}+50x-50y-18xy+77=0$
Equation of the hyperbola:
$7{x}^{2}+7{y}^{2}+50x-50y-18xy+77=0$

#### Question 2:

Find the equation of the hyperbola whose
(i) focus is (0, 3), directrix is x + y − 1 = 0 and eccentricity = 2
(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity = $\sqrt{3}$
(iv) focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2
(v) focus is (a, 0), directrix is 2xy + a = 0 and eccentricity = $\frac{4}{3}$
(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2.

(i) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
$⇒$$\sqrt{\left(x-0{\right)}^{2}+\left(y-3{\right)}^{2}}=2\left(\frac{x+y-1}{\sqrt{2}}\right)$
Squaring both the sides:
$\left(x-0{\right)}^{2}+\left(y-3{\right)}^{2}=4{\left(\frac{x+y-1}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+9-6y=2\left({x}^{2}+{y}^{2}+1+2xy-2y-2x\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+4xy+2y-4x-7=0$
∴ Equation of the hyperbola = ${x}^{2}+{y}^{2}+4xy+2y-4x-7=0$

(ii) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$⇒$$\sqrt{\left(x-1{\right)}^{2}+\left(y-1{\right)}^{2}}=2\left(\frac{3x+4y+8}{5}\right)$
Squaring both the sides:
$\left(x-1{\right)}^{2}+\left(y-1{\right)}^{2}=4{\left(\frac{3x+4y+8}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+1-2x+{y}^{2}+1-2y=\frac{4}{25}\left(9{x}^{2}+16{y}^{2}+64+24xy+64y+48x\right)\phantom{\rule{0ex}{0ex}}⇒25{x}^{2}+25-50x+25{y}^{2}+25-50y=36{x}^{2}+64{y}^{2}+256+96xy+256y+192x\phantom{\rule{0ex}{0ex}}⇒11{x}^{2}+39{y}^{2}+96xy+306y+242x+206=0$
∴ Equation of the hyperbola = $11{x}^{2}+39{y}^{2}+96xy+306y+242x+206=0$

(iii) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$⇒$$\sqrt{\left(x-1{\right)}^{2}+\left(y-1{\right)}^{2}}=\sqrt{3}\left(\frac{2x+y-1}{\sqrt{5}}\right)$
Squaring both the sides:
$\left(x-1{\right)}^{2}+\left(y-1{\right)}^{2}=3{\left(\frac{2x+y-1}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+1-2x+{y}^{2}+1-2y=\frac{3}{5}\left(4{x}^{2}+{y}^{2}+1+4xy-2y-4x\right)\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+5-10x+5{y}^{2}+5-10y=12{x}^{2}+3{y}^{2}+3+12xy-6y-12x\phantom{\rule{0ex}{0ex}}⇒7{x}^{2}-2{y}^{2}+12xy+4y-2x-7=0$
∴ Equation of the hyperbola = $7{x}^{2}-2{y}^{2}+12xy+4y-2x-7=0$

(iv) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
= ePM
$⇒$$\sqrt{\left(x-2{\right)}^{2}+\left(y+1{\right)}^{2}}=2\left(\frac{2x+3y-1}{\sqrt{13}}\right)$
Squaring both the sides:
$\left(x-2{\right)}^{2}+\left(y+1{\right)}^{2}=4{\left(\frac{2x+3y-1}{13}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4-4x+{y}^{2}+1+2y=\frac{4}{13}\left(4{x}^{2}+9{y}^{2}+1+12xy-6y-4x\right)\phantom{\rule{0ex}{0ex}}⇒13{x}^{2}+52-52x+13{y}^{2}+13+26y=16{x}^{2}+36{y}^{2}+4+48xy-24y-16x\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}+23{y}^{2}+48xy-50y+36x-61=0$
∴ Equation of the hyperbola = $3{x}^{2}+23{y}^{2}+48xy-50y+36x-61=0$

(v) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.

By definition:
SP = ePM
$⇒$$\sqrt{\left(x-a{\right)}^{2}+\left(y-0{\right)}^{2}}=\frac{4}{3}\left(\frac{2x-y+a}{\sqrt{5}}\right)$
Squaring both the sides:
$\left(x-a{\right)}^{2}+\left(y{\right)}^{2}=\frac{16}{9}{\left(\frac{2x-y+a}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2ax+{a}^{2}+{y}^{2}=\frac{16}{45}\left(4{x}^{2}+{y}^{2}+{a}^{2}-4xy-2ya+4xa\right)\phantom{\rule{0ex}{0ex}}⇒45{x}^{2}-90ax+45{a}^{2}+45{y}^{2}=64{x}^{2}+16{y}^{2}+16{a}^{2}-64xy-32ay+64ax\phantom{\rule{0ex}{0ex}}⇒19{x}^{2}-29{y}^{2}-64xy-32ay+154ax-29{a}^{2}=0$
∴ Equation of the hyperbola = $19{x}^{2}-29{y}^{2}-64xy-32ay+154ax-29{a}^{2}=0$

(vi) Let be the focus and $P\left(x,y\right)$ be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

$⇒$$\sqrt{\left(x-2{\right)}^{2}+\left(y-2{\right)}^{2}}=2\left(\frac{x+y-9}{\sqrt{2}}\right)$
Squaring both the sides:
$\left(x-2{\right)}^{2}+\left(y-2{\right)}^{2}=4{\left(\frac{x+y-9}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x+4+{y}^{2}-4y+4=2\left({x}^{2}+{y}^{2}+81+2xy-18y-18x\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x+4+{y}^{2}-4y+4=2{x}^{2}+2{y}^{2}+162+4xy-36y-36x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+4xy-32y-32x+154=0$
∴ Equation of the hyperbola = ${x}^{2}+{y}^{2}+4xy-32y-32x+154=0$

#### Question 3:

Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola
(i) 9x2 − 16y2 = 144
(ii) 16x2 − 9y2 = −144
(iii) 4x2 − 3y2 = 36
(iv) 3x2y2 = 4
(v) 2x2 − 3y2 = 5.

(i) Equation of the hyperbola:
$9{x}^{2}-16{y}^{2}=144$
This can be rewritten in the following way:
$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola, where .

$⇒{b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒9=16\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{9}{16}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{25}{16}\phantom{\rule{0ex}{0ex}}⇒e=\frac{5}{4}$
Coordinates of the foci are given by $\left(±ae,0\right)$, i.e. $\left(±5,0\right)$.
Equation of directrices:
$x=±\frac{a}{e}$
$⇒x=±\frac{4}{5}{4}}\phantom{\rule{0ex}{0ex}}⇒5x±16=0$
Length of the latus rectum of the hyperbola is $\frac{2{b}^{2}}{a}$.
Length of the latus rectum = $\frac{2×9}{4}=\frac{9}{2}$

(ii) Equation of the hyperbola:
$16{x}^{2}-9{y}^{2}=-144$
This can be rewritten in the following way:
$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=-1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola, where .

$⇒{a}^{2}={b}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒9=16\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{9}{16}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{25}{16}\phantom{\rule{0ex}{0ex}}⇒e=\frac{5}{4}$
Coordinates of foci are given by $\left(0,±ae\right)$, i.e. $\left(0,±5\right)$.
Equation of the directrices:
$y=±\frac{a}{e}$
$⇒y=±\frac{4}{5}{4}}\phantom{\rule{0ex}{0ex}}⇒5y±16=0$
Length of the latus rectum of the hyperbola = $\frac{2{a}^{2}}{b}$
Length of the latus rectum = $\frac{2×9}{4}=\frac{9}{2}$

(iii) Equation of the hyperbola:
$4{x}^{2}-3{y}^{2}=36$
This can be rewritten in the following way:
$\frac{4{x}^{2}}{36}-\frac{3{y}^{2}}{36}=1\phantom{\rule{0ex}{0ex}}\frac{{x}^{2}}{9}-\frac{{y}^{2}}{12}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola, where .

$⇒{b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒12=9\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{4}{3}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{7}{3}\phantom{\rule{0ex}{0ex}}⇒e=\sqrt{\frac{7}{3}}$
Coordinates of the foci are given by $\left(±ae,0\right)$, i.e. $\left(±\sqrt{21},0\right)$.
Equation of the directrices:
$x=±\frac{a}{e}$
$⇒x=±\frac{3}{\sqrt{7}{3}}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}x±3\sqrt{3}=0$
Length of the latus rectum of the hyperbola is $\frac{2{b}^{2}}{a}$.
$⇒\frac{2×12}{3}=8$

(iv) Equation of the hyperbola:
$3{x}^{2}-{y}^{2}=4$
This can be rewritten in the following way:
$\frac{3{x}^{2}}{4}-\frac{{y}^{2}}{4}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{4}{3}}-\frac{{y}^{2}}{4}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola, where .

$⇒{b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒4=\frac{4}{3}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=3\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=4\phantom{\rule{0ex}{0ex}}⇒e=2$
Coordinates of the foci are given by $\left(±ae,0\right)$, i.e. $\left(±\frac{4\sqrt{3}}{3},0\right)$.
Equation of the directrices:
$x=±\frac{a}{e}$
$x=±\frac{\sqrt{4}{3}}}{2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x±1=0$
Length of the latus rectum of the hyperbola = $\frac{2{b}^{2}}{a}$
$⇒\frac{2×4}{\sqrt{4}{3}}}=4\sqrt{3}$

(v) Equation of the hyperbola:
$2{x}^{2}-3{y}^{2}=5$
This can be rewritten in the following manner:
$\frac{2{x}^{2}}{5}-\frac{3{y}^{2}}{5}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{5}{2}}-\frac{{y}^{2}}{5}{3}}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola, where .

$⇒{b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{5}{3}=\frac{5}{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒e=\sqrt{\frac{5}{3}}$
Coordinates of the foci are given by $\left(±ae,0\right)$, i.e. $\left(±\frac{5\sqrt{6}}{6},0\right)$.
Equation of the directrices:
$x=±\frac{a}{e}$

Length of the latus rectum of the hyperbola is $\frac{2{b}^{2}}{a}$.

$⇒\frac{2×\left(5}{3}\right)}{\sqrt{5}{2}}}=\frac{10}{3}\sqrt{\frac{2}{5}}\phantom{\rule{0ex}{0ex}}$

#### Question 4:

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 − 36y2 = 225.

Equation of the hyperbola:
$25{x}^{2}-36{y}^{2}=225$
This equation can be rewritten in the following way:
$\frac{25{x}^{2}}{225}-\frac{36{y}^{2}}{225}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{9}-\frac{{y}^{2}}{225}{36}}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of the hyperbola, where .
Length of the transverse = $2a=2×3=6$
Length of the conjugate axis = $2b=2×\frac{15}{6}=5$

Eccentricity of the hyperbola is calculated using ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$.

$⇒\frac{225}{36}=9\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{25}{36}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{61}{36}\phantom{\rule{0ex}{0ex}}⇒e=\frac{\sqrt{61}}{6}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Length of the latus rectum =$\frac{2{b}^{2}}{a}=\frac{2×\left(225}{36}\right)}{3}=\frac{25}{6}$
The coordinates of the foci are given by $\left(±ae,0\right)$.
$⇒\left(±\frac{\sqrt{61}}{2},0\right)$

#### Question 5:

Find the centre, eccentricity, foci and directrices of the hyperbola
(i) 16x2 − 9y2 + 32x + 36y − 164 = 0
(ii) x2y2 + 4x = 0
(iii) x2 − 3y2 − 2x = 8.

(i) The equation $16{x}^{2}-9{y}^{2}+32x+36y-164=0$ can be simplified in the following way:
$16\left({x}^{2}+2x\right)-9\left({y}^{2}-4y\right)=164\phantom{\rule{0ex}{0ex}}⇒16\left({x}^{2}+2x+1\right)-9\left({y}^{2}-4y+4\right)=164+16-36\phantom{\rule{0ex}{0ex}}⇒16\left(x+1{\right)}^{2}-9{\left(y-2\right)}^{2}=144\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x+1{\right)}^{2}}{9}-\frac{\left(y-2{\right)}^{2}}{16}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the centre is $\left(-1,2\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{{a}^{2}+{b}^{2}}}{a}=\frac{\sqrt{9+16}}{3}=\frac{5}{3}$
Foci = $\left(-1±5,2\right)=\left(-6,2\right),\left(4,2\right)$

Equation of the directrices:

(ii) The equation ${x}^{2}-{y}^{2}+4x=0$ can be simplified in the following manner:
$\left({x}^{2}+4x+4\right)-{y}^{2}=4\phantom{\rule{0ex}{0ex}}⇒{\left(x+2\right)}^{2}-{y}^{2}=4\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x+2{\right)}^{2}}{4}-\frac{{y}^{2}}{1}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the centre is $\left(-2,0\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{{a}^{2}+{b}^{2}}}{a}=\frac{\sqrt{4+1}}{2}=\frac{\sqrt{5}}{2}$
Foci = $\left(-2±\sqrt{5},0\right)$
Equation of the directrices:
$x+2=±\frac{a}{e}\phantom{\rule{0ex}{0ex}}⇒x+2=±\sqrt{5}\phantom{\rule{0ex}{0ex}}$

(iii) The equation ${x}^{2}-3{y}^{2}-2x=8$ can be simplified in the following manner:
$\left({x}^{2}-2x+1\right)-3{y}^{2}=8+1\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}-3{y}^{2}=9\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x-1{\right)}^{2}}{9}-\frac{{y}^{2}}{3}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the centre is $\left(1,0\right)$.
Eccentricity of the hyperbola = $\frac{\sqrt{{a}^{2}+{b}^{2}}}{a}=\frac{\sqrt{9+3}}{3}=\frac{2\sqrt{3}}{3}$
Foci = $\left(1±2\sqrt{3},0\right)$
Equation of the directrices:
$x-1=±\frac{a}{e}\phantom{\rule{0ex}{0ex}}⇒x=1±\frac{2\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}$

#### Question 6:

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distance between the foci = 16 and eccentricity = $\sqrt{2}$
(ii) conjugate axis is 5 and the distance between foci = 13
(iii) conjugate axis is 7 and passes through the point (3, −2).

(i) The distance between the foci is $2ae$.

$\therefore 2ae=16\phantom{\rule{0ex}{0ex}}⇒ae=8$

$e=\sqrt{2}$

Therefore, the standard form of the hyperbola is given below:
$\frac{{x}^{2}}{32}-\frac{{y}^{2}}{32}=1\phantom{\rule{0ex}{0ex}}{x}^{2}-{y}^{2}=32$

(ii) The distance between the foci is $2ae$.

Length of the conjugate axis, $2b=5$
$⇒b=\frac{5}{2}$

Therefore, the standard form of the hyperbola is $\frac{{x}^{2}}{36}-\frac{4{y}^{2}}{25}=1$.

(iii) Length of the conjugate axis, $2b=7$
$⇒b=\frac{7}{2}$
Let the equation of the hyperbola be $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$.
It passes through $\left(3,-2\right)$.

Therefore, the standard form of the hyperbola is $\frac{65{x}^{2}}{441}-\frac{4{y}^{2}}{49}=1$.

#### Question 7:

Find the equation of the hyperbola whose
(i) foci are (6, 4) and (−4, 4) and eccentricity is 2.
(ii) vertices are (−8, −1) and (16, −1) and focus is (17, −1)
(iii) foci are (4, 2) and (8, 2) and eccentricity is 2.
(iv) vertices are at (0 ± 7) and foci at $\left(0,±\frac{28}{3}\right)$.
(v) vertices are at (± 6, 0) and one of the directrices is x = 4. [NCERT EXEMPLAR]
(vi) foci at (± 2, 0) and eccentricity is 3/2.   [NCERT EXEMPLAR]

(i) The centre of the hyperbola is the midpoint of the line joining  the two focii.
So, the coordinates of the centre are .
Let 2a and 2b be the length of the transverse and conjugate axis and let e be the eccentricity.

$⇒\frac{{\left(x-1\right)}^{2}}{{a}^{2}}-\frac{{\left(y-4\right)}^{2}}{{b}^{2}}=1$
Distance between the two focii = 2ae

$2ae=\sqrt{{\left(6+4\right)}^{2}+{\left(4-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒2ae=10\phantom{\rule{0ex}{0ex}}⇒ae=5\phantom{\rule{0ex}{0ex}}⇒a=\frac{5}{2}$

Equation of the hyperbola is given below:
$\frac{4{\left(x-1\right)}^{2}}{25}-\frac{4{\left(y-4\right)}^{2}}{75}=1\phantom{\rule{0ex}{0ex}}⇒\frac{4\left({x}^{2}-2x+1\right)}{25}-\frac{\left({y}^{2}-8y+16\right)}{75}=1\phantom{\rule{0ex}{0ex}}⇒\frac{\left(4{x}^{2}-8x+4\right)}{25}-\frac{\left(4{y}^{2}-32y+64\right)}{75}=1\phantom{\rule{0ex}{0ex}}⇒3\left(4{x}^{2}-8x+4\right)-\left(4{y}^{2}-32y+64\right)=75\phantom{\rule{0ex}{0ex}}⇒12{x}^{2}-24x+12-4{y}^{2}+32y-64=75\phantom{\rule{0ex}{0ex}}⇒12{x}^{2}-24x-4{y}^{2}+32y-127=0\phantom{\rule{0ex}{0ex}}$

(ii) The centre of the hyperbola is given below:
$\left(\frac{-8+16}{2},\frac{-1-1}{2}\right)=\left(4,-1\right)$
If the other focus is $S\text{'}\left(m,n\right)$, then it is calculated in the following way:

Thus, the other focus is $\left(-9,-1\right)$.

Distance between the foci:
$2c=\sqrt{{\left(17+9\right)}^{2}+{\left(-1+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒2c=26\phantom{\rule{0ex}{0ex}}⇒c=13$

Equation of the hyperbola is given below:
$\frac{{\left(x-4\right)}^{2}}{144}-\frac{{\left(y+1\right)}^{2}}{25}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}-8x+16}{144}-\frac{\left({y}^{2}+2y+1\right)}{25}=1\phantom{\rule{0ex}{0ex}}⇒25{x}^{2}-200x+400-\left(144{y}^{2}+288y+144\right)=3600\phantom{\rule{0ex}{0ex}}⇒25{x}^{2}-200x-144{y}^{2}-288y-3344=0$

(iii) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are .
Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

$⇒\frac{{\left(x-6\right)}^{2}}{{a}^{2}}-\frac{{\left(y-2\right)}^{2}}{{b}^{2}}=1$
Distance between the two focii = 2ae

$2ae=\sqrt{{\left(4-8\right)}^{2}+{\left(2-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒2ae=4\phantom{\rule{0ex}{0ex}}⇒ae=2\phantom{\rule{0ex}{0ex}}⇒a=1$

Equation of the hyperbola:
$\frac{{\left(x-6\right)}^{2}}{1}-\frac{{\left(y-2\right)}^{2}}{3}=1\phantom{\rule{0ex}{0ex}}⇒\frac{\left({x}^{2}-12x+36\right)}{1}-\frac{\left({y}^{2}-4y+4\right)}{3}=1\phantom{\rule{0ex}{0ex}}⇒3\left({x}^{2}-12x+36\right)-\left({y}^{2}-4y+4\right)=3\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-36x+108-{y}^{2}+4y-4=3\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-{y}^{2}-36x+4y+101=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(iv) The Vertices of the hyperbola are $\left(0,±7\right)$.
∴ $b=7$
The foci is $\left(0,±\frac{28}{3}\right)$.
∴ $be=\frac{28}{3}$

Therefore, the equation of the hyperbola is$-\frac{9{x}^{2}}{343}+\frac{{y}^{2}}{49}=1$.

(v) The Vertices of the hyperbola are .
∴ $a=6$
a2 = 36
Now, x = 4

Now,
${\left(ae\right)}^{2}={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(6×\frac{3}{2}\right)}^{2}={6}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒81-36={b}^{2}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=45$

Therefore, the equation of the hyperbola is$\frac{{x}^{2}}{36}-\frac{{y}^{2}}{45}=1$.

(vi) The foci of the hyperbola are .
∴ $ae=2\phantom{\rule{0ex}{0ex}}⇒a=2×\frac{2}{3}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=\frac{16}{9}$
Now,
${\left(ae\right)}^{2}={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(2\right)}^{2}={\left(\frac{4}{3}\right)}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒4-\frac{16}{9}={b}^{2}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=\frac{20}{9}$

Therefore, the equation of the hyperbola is given by
$\frac{9{x}^{2}}{16}-\frac{9{y}^{2}}{20}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{4}-\frac{{y}^{2}}{5}=\frac{4}{9}$

#### Question 8:

Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\frac{3}{4}$ of the length of transverse axis.

The lengths of the conjugate axis and the transverse axis are $2b$ and $2a$, respectively.
We have:
$⇒2b=\frac{3}{4}×2a\phantom{\rule{0ex}{0ex}}⇒b=\frac{3}{4}a$
Using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:

$⇒{\left(\frac{3}{4}\right)}^{2}{a}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{9}{16}={e}^{2}-1\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{25}{16}\phantom{\rule{0ex}{0ex}}⇒e=\frac{5}{4}\phantom{\rule{0ex}{0ex}}$
Therefore, the eccentricity of the hyperbola is $\frac{5}{4}$.

#### Question 9:

Find the equation of the hyperboala whose
(i) focus is at (5, 2), vertex at (4, 2) and centre at (3, 2)
(ii) focus is at (4, 2), centre at (6, 2) and e = 2.

(i) The equation of the hyperbola with centre (x0,y0) is given by

Vertex = (a+x0, y0)

$⇒\frac{{\left(x-3\right)}^{2}}{1}-\frac{{\left(y-2\right)}^{2}}{3}=1\phantom{\rule{0ex}{0ex}}⇒3{\left(x-3\right)}^{2}-{\left(y-2\right)}^{2}=3$

(ii) The equation of the hyperbola with centre (x0,y0) is given by

$⇒\frac{{\left(x-6\right)}^{2}}{1}-\frac{{\left(y-2\right)}^{2}}{3}=1\phantom{\rule{0ex}{0ex}}⇒3{\left(x-6\right)}^{2}-{\left(y-2\right)}^{2}=3$

#### Question 10:

If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP2.

Equation of the hyperbola:
$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$

If the axes of the hyperbola are equal, then $a=b$.
Then, equation of the hyperbola becomes ${x}^{2}-{y}^{2}={a}^{2}$.

Thus, the centre $C\left(0,0\right)$ and the focus are given by $S\left(\sqrt{2}a,0\right)$ and $S\text{'}\left(-\sqrt{2}a,0\right)$, respectively.
Let $P\left(\alpha ,\beta \right)$ be any point on the parabola.
So, it will satisfy the equation.
${\alpha }^{2}-{\beta }^{2}={a}^{2}$

$S\text{'}{P}^{2}={\left(-\sqrt{2}a-\alpha \right)}^{2}+{\beta }^{2}\phantom{\rule{0ex}{0ex}}=2{a}^{2}+{\alpha }^{2}+2\sqrt{2}a\alpha +{\beta }^{2}$

∴ $SP.S\text{'}P=C{P}^{2}$

#### Question 11:

In each of the following find the equations of the hyperbola satisfying the given conditions:
(i) vertices (± 2, 0), foci (± 3, 0)
(ii) vertices (0, ± 5), foci (0, ± 8)
(iii) vertices (0, ± 3), foci (0, ± 5)
(iv) foci (± 5, 0), transverse axis = 8
(v) foci (0, ± 13), conjugate axis = 24
(vi) foci (± $3\sqrt{5}$, 0), the latus-rectum = 8
(vii) foci (± 4, 0), the latus-rectum = 12
(viii) vertices (± 7, 0), $e=\frac{4}{3}$
(ix) foci (0, ± $\sqrt{10}$), passing through (2, 3)
(x) foci (0, ± 12), latus-rectum = 36

(i) The vertices of the hyperbola are $\left(±2,0\right)$ and the foci are $\left(±3,0\right)$.
Thus, the value of $a=2$ and $ae=3$.
Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{b}^{2}=9-4\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=5\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is $\frac{{x}^{2}}{4}-\frac{{y}^{2}}{5}=1$.

(ii) The vertices of the hyperbola are $\left(0,±5\right)$ and the foci are $\left(0,±8\right)$
Thus, the value of $a=5$ and $ae=8$.
Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{b}^{2}=64-25\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=39\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is $-\frac{{x}^{2}}{39}+\frac{{y}^{2}}{25}=1$.

(iii) The vertices of the hyperbola are $\left(0,±3\right)$ and the foci are $\left(0,±5\right)$.
Thus, the value of $a=3$ and $ae=5$.
Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{b}^{2}=25-9\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=16\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is  $-\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$.

(iv) The foci of  the hyperbola  are $\left(±5,0\right)$ and the transverse axis is 8.
Thus, the value of  $ae=5$ and 2a = 8.
$⇒a=4$
Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{b}^{2}=25-16\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=9\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$

(v) The foci of the hyperbola are $\left(0,±13\right)$ and the conjugate axis is 24.
Thus, the value of  $ae=13$ and 2b = 24.
b = 12

Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{a}^{2}=169-144\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=25\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is$-\frac{{x}^{2}}{144}+\frac{{y}^{2}}{25}=1$

(vi) The foci of the hyperbola are $\left(±3\sqrt{5},0\right)$ and the latus rectum is 8.
Thus, the value of  $ae=3\sqrt{5}$

Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:

Since negative value is not possible, it is equal to 20.
Thus, the equation of the hyperbola is$\frac{{x}^{2}}{25}-\frac{{y}^{2}}{20}=1$

(vii) The foci of the hyperbola are $\left(±4,0\right)$ and the latus rectum is 12.
Thus, the value of  $ae=4$

Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:

Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola is$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{12}=1$

(viii) The vertices of hyperbola are $\left(±7,0\right)$ and eccentricity is $\frac{4}{3}$
Thus, the value of $a=7$.
Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
$⇒{b}^{2}=49\left(\frac{16}{9}-1\right)\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=49×\frac{7}{9}=\frac{343}{9}\phantom{\rule{0ex}{0ex}}$
Thus, the equation of the hyperbola is$\frac{{x}^{2}}{49}-\frac{9{y}^{2}}{343}=1$.

(ix) The foci of hyperbola are $\left(0,±\sqrt{10}\right)$ that pass through $\left(2,3\right)$.

Let the equation of the hyperbola be $\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1$.
It passes through $\left(2,3\right)$.
$⇒\frac{{3}^{2}}{{a}^{2}}-\frac{{2}^{2}}{10-{a}^{2}}=1\phantom{\rule{0ex}{0ex}}⇒90-9{a}^{2}-4{a}^{2}=10{a}^{2}-{a}^{4}\phantom{\rule{0ex}{0ex}}⇒{a}^{4}-23{a}^{2}+90=0\phantom{\rule{0ex}{0ex}}⇒\left({a}^{2}-18\right)\left({a}^{2}-5\right)=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=18,5$
Now,
If we neglect the negative value,  then b2 = 5.
Thus, the equation of the hyperbola is$\frac{{y}^{2}}{5}-\frac{{x}^{2}}{5}=1$.

(x) The foci of the hyperbola are $\left(0,±12\right)$ and the latus rectum is 36.
Thus, the value of  $ae=12$.

Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:

Thus, the equation of the hyperbola is$\frac{{y}^{2}}{36}-\frac{{x}^{2}}{108}=1$.

#### Question 12:

If the distance between the foci of a hyperbola is 16 and its ecentricity is $\sqrt{2}$, then obtain its equation.

We have
$2ae=16\phantom{\rule{0ex}{0ex}}⇒ae=8\phantom{\rule{0ex}{0ex}}⇒a=\frac{8}{\sqrt{2}}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=32$
Now,
${\left(ae\right)}^{2}={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(8\right)}^{2}=32+{b}^{2}\phantom{\rule{0ex}{0ex}}⇒64-32={b}^{2}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=32$

Therefore, the equation of the hyperbola is given by
$\frac{{x}^{2}}{32}-\frac{{y}^{2}}{32}=1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-{y}^{2}=32$

#### Question 13:

Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.

Let the point be P(xy)
$\therefore \left[\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}}\right]-\left[\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\right]=2\phantom{\rule{0ex}{0ex}}⇒{\left[\sqrt{{\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}}\right]}^{2}={\left[2+\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\right]}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-4\right)}^{2}+{y}^{2}=4+{\left(x+4\right)}^{2}+{y}^{2}+4\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\left(x-4\right)}^{2}-{\left(x+4\right)}^{2}=4+4\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}$â€‹
$⇒-16x=4+4\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒-16x-4=4\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒-4\left(4x+1\right)=4\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒-\left(4x+1\right)=\sqrt{{\left(x+4\right)}^{2}+{\left(y-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+8x+1={x}^{2}+8x+16+{y}^{2}\phantom{\rule{0ex}{0ex}}⇒15{x}^{2}-{y}^{2}=15\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{1}-\frac{{y}^{2}}{15}=1$
Which is the equation of a hyperbola.

#### Question 1:

Write the eccentricity of the hyperbola 9x2 − 16y2 = 144.

Equation of the hyperbola:
..... (1)

Equation (1) can be rewritten in the following way:

$⇒\frac{{x}^{2}}{{4}^{2}}-\frac{{y}^{2}}{{3}^{2}}=1$
This becomes the standard equation of the hyperbola with its major axis $a=4$ and minor axis $b=3$.
Eccentricity,
Substituting the value of a and b, we get:

Therefore, the eccentricity is $\frac{5}{4}$.

#### Question 2:

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

The lengths of the latus rectum and the transverse axis are $\frac{2{b}^{2}}{a}$ and $2a$, respectively. â€‹
According to the given statement, length of the latus rectum is half of its trasverse axis.

Eccentricity,
Substituting the value ${b}^{2}=\frac{{a}^{2}}{2}$, we get:

Therefore, the eccentricity is $\sqrt{\frac{3}{2}}$.

#### Question 3:

Write the coordinates of the foci of the hyperbola 9x2 − 16y2 = 144.

Equation of the hyperbola:
$9{x}^{2}-16{y}^{2}=144$ ..... (1)

This equation (1) can be rewritten in the following way:
$⇒\frac{9{x}^{2}}{144}-\frac{16{y}^{2}}{144}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$

This becomes a standard form of the hyperbola with transverse axis $a=4$ and conjugate axis $b=3$.

The foci of the hyperbola are of the form $\left(ae,0\right)$ and
$\left(-ae,0\right)$.
Therefore, the foci are $\left(5,0\right)$ and $\left(-5,0\right)$.4

#### Question 4:

Write the equation of the hyperbola of eccentricity $\sqrt{2}$, if it is known that the distance between its foci is 16.

The foci of the hyperbola are of the form $\left(ae,0\right)$ and $\left(-ae,0\right)$.

Distance between the foci is 16 and eccentricity of the hyperbola is $\sqrt{2}$.

Equation of the hyperbola is given below:
$\frac{{x}^{2}}{{\left(4\sqrt{2}\right)}^{2}}-\frac{{y}^{2}}{32}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{32}-\frac{{y}^{2}}{32}=1$

#### Question 5:

If the foci of the ellipse $\frac{{x}^{2}}{16}+\frac{{y}^{2}}{{b}^{2}}=1$ and the hyperbola $\frac{{x}^{2}}{144}-\frac{{y}^{2}}{81}=\frac{1}{25}$ coincide, write the value of b2.

Equation of the hyperbola:
$\frac{{x}^{2}}{144}-\frac{{y}^{2}}{81}=\frac{1}{25}$

Simplifying the above equation, we get:
$\frac{25{x}^{2}}{144}-\frac{25{y}^{2}}{81}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{144}{25}}-\frac{{y}^{2}}{81}{25}}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{{\left(12}{5}\right)}^{2}}-\frac{{y}^{2}}{{\left(9}{5}\right)}^{2}}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
This is the standard form of hyperbola with $a=\frac{12}{5}$and $b=\frac{9}{5}$.

The foci of the hyperbola are of the form $\left(ae,0\right)$ and $\left(-ae,0\right)$.
So, foci of the hyperbola are $\left(3,0\right)$ and $\left(-3,0\right)$.
The foci of the hyperbola coincide with the foci of the ellipse.

For ellipse,
$a=4$

Using the relation$\sqrt{{a}^{2}-{b}^{2}}=ae$, we get:
$\sqrt{{4}^{2}-{b}^{2}}=3\phantom{\rule{0ex}{0ex}}⇒16-{b}^{2}=9\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=7$
Therefore, the value of ${b}^{2}$ is 7.

#### Question 6:

Write the length of the latus-rectum of the hyperbola 16x2 − 9y2 = 144.

Equation of the hyperbola:
$16{x}^{2}-9{y}^{2}=144$
This equation can be rewritten in the following way:
$\frac{16{x}^{2}}{144}-\frac{9{y}^{2}}{144}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{{3}^{2}}-\frac{{y}^{2}}{{4}^{2}}=1\phantom{\rule{0ex}{0ex}}$
This is the standard form of a hyperbola with $a=3$ and $b=4$.
Length of the latus rectum = $\frac{2{b}^{2}}{a}$
Substituting the value of and b, we get:
Length of the â€‹latus rectum  $=\frac{2×{4}^{2}}{3}=\frac{32}{3}$

#### Question 7:

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

The standard equation of hyperbola is given below:
$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\phantom{\rule{0ex}{0ex}}$

Let the latus rectum be QR, which passes through one of the focus. QR subtends a right angle at point P.

It passes through the focus $\left(ae,±y\right)$.
$\therefore \frac{{\left(ae\right)}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{y}^{2}}{{b}^{2}}=\frac{{\left(ae\right)}^{2}}{{a}^{2}}-1\phantom{\rule{0ex}{0ex}}⇒\frac{{y}^{2}}{{b}^{2}}={e}^{2}-1\phantom{\rule{0ex}{0ex}}⇒y=±b\sqrt{{e}^{2}-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now, the slope of PQ is $\frac{b\sqrt{{e}^{2}-1}}{ae+a}$ and the slope of PR is$-\frac{b\sqrt{{e}^{2}-1}}{ae+a}$.
The lines PQ and PR are perpendicular, so the product of the slope is $-1$.

$⇒\frac{b\sqrt{{e}^{2}-1}}{ae+a}×\frac{-b\sqrt{{e}^{2}-1}}{ae+a}=-1\phantom{\rule{0ex}{0ex}}⇒{b}^{2}\left({e}^{2}-1\right)={\left(ae+a\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}\left({e}^{2}-1\right)={a}^{2}{\left(e+1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left({e}^{2}-1\right)}^{2}={\left(e+1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{e}^{4}-2{e}^{2}+1={e}^{2}+2e+1\phantom{\rule{0ex}{0ex}}⇒{e}^{4}-3{e}^{2}-2e=0\phantom{\rule{0ex}{0ex}}⇒e\left({e}^{3}-3e-2\right)=0\phantom{\rule{0ex}{0ex}}⇒e\left(e-2\right)\left(e+1\right)\left(e+1\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ $e=0,-1,2$

The value of e can neither be negative nor zero.

Therefore, the value of eccentricity is 2.

#### Question 8:

Write the distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ.

We have:

a = b = 8
Distance between the directrices of  hyperbola is $\frac{2{a}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}$.

$⇒\frac{2×64}{\sqrt{64+64}}\phantom{\rule{0ex}{0ex}}=\frac{128}{8\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{16}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=8\sqrt{2}$

#### Question 9:

Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).

The vertices of the hyperbola and the foci are $\left(±a,0\right)$ and $\left(±ae,0\right)$, respectively.

Using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:

$⇒{b}^{2}={\left(ae\right)}^{2}-{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=25-9\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=16\phantom{\rule{0ex}{0ex}}⇒b=4\phantom{\rule{0ex}{0ex}}$
Therefore, the equation of hyperbola is $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1$.

#### Question 10:

If e1 and e2 are respectively the eccentricities of the ellipse $\frac{{x}^{2}}{18}+\frac{{y}^{2}}{4}=1$ and the hyperbola $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1$, then write the value of 2 e12 + e22.

The standard form of the ellipse is $\frac{{x}^{2}}{18}+\frac{{y}^{2}}{4}=1$, where .
So, the eccentricity is calculated in the following way:
$⇒{b}^{2}={a}^{2}\left(1-{{e}_{1}}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒4=18\left(1-{{e}_{1}}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{9}=1-{{e}_{1}}^{2}\phantom{\rule{0ex}{0ex}}⇒{{e}_{1}}^{2}=\frac{7}{9}$
The standard form of the hyperbola is $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1$, where .
So, the eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left({{e}_{2}}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒4=9\left({{e}_{2}}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{4}{9}={{e}_{2}}^{2}-1\phantom{\rule{0ex}{0ex}}⇒{{e}_{2}}^{2}=\frac{13}{9}$

#### Question 1:

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
(a) 16x2 − 9y2 = 144
(b) 9x2 − 16y2 = 144
(c) 25x2 − 9y2 = 225
(d) 9x2 − 25y2 = 81

(a) 16x2 − 9y2 = 144

The vertices of the hyperbola are $\left(±3,0\right)$ and foci are $\left(±5,0\right)$.
Thus, the values of a and ae are 3 and 5, respectively.

Now, using the relation ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$, we get:
${b}^{2}=25-9\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=16\phantom{\rule{0ex}{0ex}}$
Equation of the hyperbola is given below:
$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-9{y}^{2}=144$

#### Question 2:

If e1 and e2 are respectively the eccentricities of the ellipse $\frac{{x}^{2}}{18}+\frac{{y}^{2}}{4}=1$ and the hyperbola $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1$, then the relation between e1 and e2 is
(a) 3 e12 + e22 = 2
(b) e12 + 2 e22 = 3
(c) 2 e12 + e22 = 3
(d) e12 + 3 e22 = 2

(c) 2 e12 + e22 = 3
The standard form of the ellipse is $\frac{{x}^{2}}{18}+\frac{{y}^{2}}{4}=1$, where .
So, the eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left(1-{{e}_{1}}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒4=18\left(1-{{e}_{1}}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{9}=1-{{e}_{1}}^{2}\phantom{\rule{0ex}{0ex}}⇒{{e}_{1}}^{2}=\frac{7}{9}$
The standard form of the hyperbola is $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1$, where .
So, the eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left({{e}_{2}}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒4=9\left({{e}_{2}}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{4}{9}={{e}_{2}}^{2}-1\phantom{\rule{0ex}{0ex}}⇒{{e}_{2}}^{2}=\frac{13}{9}$

#### Question 3:

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is
(a) $8\sqrt{2}$
(b) $16\sqrt{2}$
(c) $4\sqrt{2}$
(d) $6\sqrt{2}$

(a) $8\sqrt{2}$

We have:

∴ a = b = 8
Distance between the directrices of the hyperbola = $\frac{2{a}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}$

#### Question 4:

The equation of the conic with focus at (1, 1) directrix along xy + 1 = 0 and eccentricity $\sqrt{2}$ is
(a) xy = 1
(b) 2xy + 4x − 4y − 1= 0
(c) x2y2 = 1
(d) 2xy − 4x + 4y + 1 = 0

(d) 2xy − 4x + 4y + 1 = 0

Let $P\left(x,y\right)$ be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.

Squaring both the sides, we get:
$\left(x-1{\right)}^{2}+\left(y+1{\right)}^{2}={\left(x-y+1\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}-2x+1+{y}^{2}+1+2y={x}^{2}+{y}^{2}+1-2xy-2y+2x\phantom{\rule{0ex}{0ex}}2xy-4x+4y+1=0\phantom{\rule{0ex}{0ex}}$

#### Question 5:

The eccentricity of the conic 9x2 − 16y2 = 144 is
(a) $\frac{5}{4}$

(b) $\frac{4}{3}$

(c) $\frac{4}{5}$

(d) $\sqrt{7}$

(a) $\frac{5}{4}$

Standard form of a hyperbola = $\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$
Here,

The eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒9=16\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{9}{16}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{25}{16}\phantom{\rule{0ex}{0ex}}⇒e=\frac{5}{4}$

#### Question 6:

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PAPB = k (k ≠ 0), then the locus of P is
(a) a hyperbola
(b) a branch of the hyperbola
(c) a parabola
(d) an ellipse

(a) a hyperbola

#### Question 7:

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is
(a)$\frac{1}{\sqrt{2}}$

(b) $\sqrt{\frac{2}{3}}$

(c) $\sqrt{\frac{3}{2}}$

(d) none of these.

(c) $\sqrt{\frac{3}{2}}$
The lengths of the latus rectum and the transverse axis are $\frac{2{b}^{2}}{a}$ and $2a$, respectively. â€‹
According to the given statement, length of the latus rectum is half of its transverse axis.

Eccentricity,
Substituting the value ${b}^{2}=\frac{{a}^{2}}{2}$ , we get:

∴ Eccentricity is $\sqrt{\frac{3}{2}}$

#### Question 8:

The eccentricity of the hyperbola x2 − 4y2 = 1 is

(a) $\frac{\sqrt{3}}{2}$

(b) $\frac{\sqrt{5}}{2}$

(c) $\frac{2}{\sqrt{3}}$

(d) $\frac{2}{\sqrt{5}}$

(b) $\frac{\sqrt{5}}{2}$

The equation of the hyperbola is ${x}^{2}-4{y}^{2}=1$.
This can be rewritten in the following way:
$\frac{{x}^{2}}{1}-\frac{{y}^{2}}{1}{4}}=1$
This is the standard form of a hyperbola, where .
The value of eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{4}=\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒e=\frac{\sqrt{5}}{2}$

#### Question 9:

The difference of the focal distances of any point on the hyperbola is equal to
(a) length of the conjugate axis
(b) eccentricity
(c) length of the transverse axis
(d) Latus-rectum

(c) length of the transverse axis

Let $P\left(x,y\right)$be any point on the hyperbola, and  be the focus with coordinates $\left(±ae,0\right)$.

$S\text{'}P-SP=2a$
Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

#### Question 10:

The foci of the hyperbola 9x2 − 16y2 = 144 are
(a) (± 4, 0)
(b) (0, ± 4)
(c) (± 5, 0)
(d) (0, ± 5)

(c) (± 5, 0)

The equation of the hyperbola is given below:
$9{x}^{2}-16{y}^{2}=144$
This equation can be rewritten in the following way:
$\frac{9{x}^{2}}{144}-\frac{16{y}^{2}}{144}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$
This is the standard equation of a hyperbola, where .

The eccentricity is calculated in the following way:
${b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒9=16\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{9}{16}={e}^{2}-1\phantom{\rule{0ex}{0ex}}⇒e=\frac{5}{4}$
Foci = $\left(±ae,0\right)=\left(±5,0\right)$

#### Question 11:

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$, then equation of the hyperbola is
(a) x2 + y2 = 32
(b) x2y2 = 16
(c) x2 + y2 = 16
(d) x2y2= 32

(d) x2y2= 32

The distance between the foci is $2ae$.

$e=\sqrt{2}$

Standard form of the hyperbola is given below:
$\frac{{x}^{2}}{32}-\frac{{y}^{2}}{32}=1\phantom{\rule{0ex}{0ex}}{x}^{2}-{y}^{2}=32$

#### Question 12:

If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 − 4y2 = 36, then
(a) e12e22 = 2
(b) 2 < e22e12 < 3
(c) e22e12 = 2
(d) e22e12 > 3

(b) 2 < e22e12 < 3
The conic â€‹$9{x}^{2}+4{y}^{2}=36$ can rewritten in the following way:
$\frac{9{x}^{2}}{36}+\frac{4{y}^{2}}{36}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of an ellipse.

The conic â€‹$9{x}^{2}-4{y}^{2}=36$ can rewritten in the following way:
$\frac{9{x}^{2}}{36}-\frac{4{y}^{2}}{36}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{4}-\frac{{y}^{2}}{9}=1\phantom{\rule{0ex}{0ex}}$
This is the standard equation of a hyperbola.

#### Question 13:

If the eccentricity of the hyperbola x2y2 sec2α = 5 is $\sqrt{3}$ times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =

(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{\mathrm{\pi }}{4}$
The hyperbola ${x}^{2}-{y}^{2}se{c}^{2}\alpha =5$ can be rewritten in the following way:
$\frac{{x}^{2}}{5}-\frac{{y}^{2}}{5{\mathrm{cos}}^{2}\alpha }=1$

This is the standard form of a hyperbola, where .

The ellipse ${x}^{2}se{c}^{2}\alpha +{y}^{2}=25$ can be rewritten in the following way:
$\frac{{x}^{2}}{25{\mathrm{cos}}^{2}\alpha }+\frac{{y}^{2}}{25}=1$

This is the standard form of an ellipse, where .

According to the question,

${\mathrm{cos}}^{2}\alpha +1=3\left({\mathrm{sin}}^{2}\alpha \right)\phantom{\rule{0ex}{0ex}}⇒2=4{\mathrm{sin}}^{2}\alpha \phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\alpha =\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\alpha =\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 14:

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

(a) $\frac{\left(x-1{\right)}^{2}}{25/4}-\frac{\left(y-4{\right)}^{2}}{75/4}=1$

(b) $\frac{\left(x+1{\right)}^{2}}{25/4}-\frac{\left(y+4{\right)}^{2}}{75/4}=1$

(c) $\frac{\left(x-1{\right)}^{2}}{75/4}-\frac{\left(y-4{\right)}^{2}}{25/4}=1$

(d) none of these

(a) $\frac{\left(x-1{\right)}^{2}}{25/4}-\frac{\left(y-4{\right)}^{2}}{75/4}=1$

The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are
Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

$⇒\frac{{\left(x-1\right)}^{2}}{{a}^{2}}-\frac{{\left(y-4\right)}^{2}}{{b}^{2}}=1$

Now, distance between the two foci = 2ae

$2ae=\sqrt{{\left(6+4\right)}^{2}+{\left(4-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒2ae=10\phantom{\rule{0ex}{0ex}}⇒ae=5\phantom{\rule{0ex}{0ex}}⇒a=\frac{5}{2}$

Equation of the hyperbola is given below:
$\frac{{\left(x-1\right)}^{2}}{25/4}-\frac{{\left(y-4\right)}^{2}}{75/4}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 15:

The length of the straight line x − 3y = 1 intercepted by the hyperbola x2 − 4y2 = 1 is

(a) $\frac{6}{\sqrt{5}}$

(b) $3\sqrt{\frac{2}{5}}$

(c) $6\sqrt{\frac{2}{5}}$

(d) none of these

(c) $6\sqrt{\frac{2}{5}}$

The point of intersection of $x-3y=1$and the hyperbola ${x}^{2}-4{y}^{2}=1$ is calculated in the following way:

If $y=0$, then $x=1$.
If $y=-\frac{6}{5}$, then $x=1+3×\left(-\frac{6}{5}\right)=-\frac{13}{5}$.

So, the points are $\left(1,0\right)$ and $\left(-\frac{13}{5},-\frac{6}{5}\right)$.

∴ Length = $\sqrt{{\left(1+\frac{13}{5}\right)}^{2}+{\left(0+\frac{6}{5}\right)}^{2}}=6\sqrt{\frac{2}{5}}$

#### Question 16:

The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is
(a) 16/3
(b) 32/3
(c) 8/3
(d) 4/3

(b) 32/3

The standard form of the hyperbola $16{x}^{2}-9{y}^{2}=144$ is $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1$.
Here,
Latus rectum of the hyperbola = $\frac{2{b}^{2}}{a}=\frac{2×16}{3}=\frac{32}{3}$

#### Question 17:

The foci of the hyperbola 2x2 − 3y2 = 5 are
(a) $\left(±5/\sqrt{6},0\right)$
(b) (± 5/6, 0)
(c) $\left(±\sqrt{5}/6,0\right)$
(d) none of these

(a) $\left(±5/\sqrt{6},0\right)$

The given equation of hyperbola is $2{x}^{2}-3{y}^{2}=5$. It can be rewritten in the following way:
$\frac{2{x}^{2}}{5}-\frac{3{y}^{2}}{5}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{\frac{5}{2}}-\frac{{y}^{2}}{\frac{5}{3}}=1\phantom{\rule{0ex}{0ex}}$

This is the standard equation of a parabola, where .
The eccentricity can be calculated in the following way:
${b}^{2}={a}^{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{5}{3}=\frac{5}{2}\left({e}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2}-1=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒{e}^{2}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒e=\sqrt{\frac{5}{3}}$
Coordinates of the foci =

#### Question 18:

The eccentricity the hyperbola is
(a) $\sqrt{2}$
(b) $\sqrt{3}$
(c) $2\sqrt{3}$
(d) $3\sqrt{2}$

(a) $\sqrt{2}$
We eliminate t in $x=\frac{a}{2}\left(t+\frac{1}{t}\right),y=\frac{a}{2}\left(t-\frac{1}{t}\right)$.
On squaring both the sides in the equations $x=\frac{a}{2}\left(t+\frac{1}{t}\right),y=\frac{a}{2}\left(t-\frac{1}{t}\right)$, we get:

From (1) and (2), we get:
$\frac{4{x}^{2}}{{a}^{2}}-\frac{4{y}^{2}}{{a}^{2}}=4\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{a}^{2}}=1\phantom{\rule{0ex}{0ex}}$

This is the standard equation of a hyperbola, where ${a}^{2}={b}^{2}$.

Eccentricity of the hyperbola, $e=\frac{\sqrt{{a}^{2}+{a}^{2}}}{a}=\sqrt{2}$

#### Question 19:

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
(a) 3 (x − 6)2 − (y −2)2 = 3
(b) (x − 6)2 − 3 (y − 2)2 = 1
(c) (x − 6)2 − 2 (y −2)2 = 1
(d) 2 (x − 6)2 − (y − 2)2 = 1

(a) 3 (x − 6)2 − (y −2)2 = 3

The equation of the hyperbola with centre (x0,y0) is given by

#### Question 20:

The locus of the point of intersection of the lines is a hyperbola of eccentricity
(a) 1
(b) 2
(c) 3
(d) 4

$3\lambda {x}^{2}-\lambda {y}^{2}=48\lambda \phantom{\rule{0ex}{0ex}}⇒\frac{3\lambda {x}^{2}}{48\lambda }-\frac{\lambda {y}^{2}}{48\lambda }=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{16}-\frac{{y}^{2}}{48}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$