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Page No 26.22:

Question 1:

Find the equation of the ellipse whose focus is (1, −2), the directrix 3x − 2y + 5 = 0 and eccentricity equal to 1/2.

Answer:



Let S(1,-2) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directix.Then by the definition of an ellipse, we have:SP=e.PM, where e=12SP2 = e2.PM2(x-1)2 + (y+2)2 =122.3x-2y+5(3)2+(-2)22x2+1-2x+y2+4+4y= 14.9x2+4y2+25-12xy-20y+30x1352(x2+1-2x+y2+4+4y) = 9x2+4y2+25-12xy-20y+30x52x2+52-104x+52y2+208+208y= 9x2+4y2+25-12xy-20y+30x43x2+48y2-134x+228y+12xy+235=0This is the equation of the required ellipse.

Page No 26.22:

Question 2:

Find the equation of the ellipse in the following cases:
(i) focus is (0, 1), directrix is x + y = 0 and e = 12
(ii) focus is (−1, 1), directrix is xy + 3 = 0 and e = 12
(iii) focus is (−2, 3), directrix is 2x + 3y + 4 = 0 and e = 45
(iv) focus is (1, 2), directrix is 3x + 4y − 5 = 0 and e = 12.

Answer:

(i)



Let S(0,1) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PMSP=12×PM2SP=PM4SP2=PM24x2+y-12=x+y12+1224x2+y2+1-2y=x2+y2+2xy28x2+8y2+8-16y=x2+y2+2xy7x2+7y2-2xy-16y+8=0This is the required equation of the ellipse.

(ii)

Let S(-1,1) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PMSP=12×PM2SP=PM4SP2=PM24x+12+y-12=x-y+312+-1224x2+1+2x+y2+1-2y=x2+y2+9-2xy-6y+6x28x2+8+16x+8y2+8-16y=x2+y2+9-2xy-6y+6x7x2+7y2+2xy-10y+10x+7=0This is the required equation of the ellipse.

(iii)


Let S(-2,3) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have: SP=e×PMSP=45×PM54SP=PM2516SP2=PM22516x+22+y-32=2x+3y+422+3222516x2+4+4x+y2+9-6y=4x2+9y2+16+12xy+24y+16x13325x2+4+4x+y2+9-6y=164x2+9y2+16+12xy+24y+16x325x2+1300+1300x+325y2+2925-1950y=64x2+144y2+256+192xy+384y+256x261x2+181y2+1044x-2309y-192xy+3969=0This is the required equation of the ellipse.

(iv)

 Let S(1,2) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PMSP=12×PM2SP=PM4SP2=PM24x-12+y-22=3x+4y-532+4224x2+1-2x+y2+4-4y=9x2+16y2+25+24xy-40y+30x25100x2+100-200x+100y2+400-400y=9x2+16y2+25+24xy-40y-30x91x2+84y2-24xy-360y-170x+475=0This is the required equation of the ellipse.

Page No 26.22:

Question 3:

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) 4x2 + 9y2 = 1
(ii) 5x2 + 4y2 = 1
(iii) 4x2 + 3y2 = 1
(iv) 25x2 + 16y2 = 1600.
(v) 9x2 + 25y2 = 225

Answer:

(i)  4x2+9y2=1x214+y219=1This is of the form x2a2+y2b2=1, where a2=14 and b2=19, i.e. a=12and b=13.Clearly a>bNow, e=1-b2a2e=1-1914e=1-49e=53Coordinates of the foci=±ae, 0=±56,oLength of the latus rectum= 2b2a                                             =2×1912                                             =49
(ii)  5x2+4y2=1x215+y214=1This is of the form x2a2+y2b2=1, where a2=15 and b2=14, i.e. a=15and b=12. Clearly b>aNow, e=1-a2b2e=1-1514e=1-45e=15Coordinates of the foci=0,±be= 0,±125Length of the latus rectum= 2a2b                                            =2×1512                                           =45

(iii) 4x2+3y2=1x214+y213=1This is of the form x2a2+y2b2=1, where a2=14 and b2=13, i.e. a=12and b=13. Clearly, b>aNow, e=1-a2b2e=1-1413e=1-34e=12Coordinates of the foci=0,±be= 0,±123Length of the latus rectum= 2a2b                                              =2×1413                                              =32

(iv)  25x2+16y2=1600x264+y2100=1This is of the form x2a2+y2b2=1, where a2=64 and b2=100, i.e. a=8 and b=10.Clearly, b>aNow, e=1-a2b2e=1-64100e=36100e=610 or 35Coordinates of the foci= 0,±6Length of latus rectum= 2a2b                                     =2×6410                                     =645
(v)  9x2+25y2=225x225+y29=1This is of the form x2a2+y2b2=1, where a2=25 and b2=9, i.e. a=5 and b=3.Clearly, a>bNow, e=1-b2a2e=1-925e=1625e=45 Coordinates of the foci=±ae, 0= ±4, 0Length of latus rectum= 2b2a                                     =2×95                                     =185

Page No 26.22:

Question 4:

Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (−3, 1) and has eccentricity 25.

Answer:

Let the equation of the ellipse be x2a2+y2b2=1          ...(1) It passes through the point -3,1.∴ 9a2+1b2=1           ... (2)Also, e=25Now, b2=a21-e2b2=a21-25b2=3a25Substituting the value of b2 in eq. (2), we get:9a2+53a2=127+53a2=1a2=323b2=3×3235or 325Substituting the values of a and b in eq. (1), we get: 3x232+5y232=13x2+5y2=32This is the required equation of the ellipse.

Page No 26.22:

Question 5:

Find the equation of the ellipse in the following cases:
(i) eccentricity e = 12 and foci (± 2, 0)
(ii) eccentricity e = 23 and length of latus rectum = 5
(iii) eccentricity e = 12 and semi-major axis = 4
(iv) eccentricity e = 12 and major axis = 12
(v) The ellipse passes through (1, 4) and (−6, 1).
(vi) Vertices (± 5, 0), foci (± 4, 0)
(vii) Vertices (0, ± 13), foci (0, ± 5)
(viii) Vertices (± 6, 0), foci (± 4, 0)
(ix) Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)
(x) Ends of major axis (0, ± 5), ends of minor axis (± 1, 0)
(xi) Length of major axis 26, foci (± 5, 0)
(xii) Length of minor axis 16 foci (0, ± 6)
(xiii) Foci (± 3, 0), a = 4

Answer:

(i)  e=12 and foci=(±2,0)Coordinates of the foci=±ae, 0We have ae=2 a×12=2a=4Now, e=1-b2a212=1-b216On squaring both sides, we get:14=16-b21664-4b2=16b2=484b2=12Now, x2a2+y2b2=1x216+y212=13x2+4y248=13x2+4y2=48This is the required equation of the ellipse.

 (ii)  e=23 and length of the latus rectum=5We have 2b2a=52b2=5ab2=5a2Now, e=1-b2a223=1-5a2a2On squaring both sides, we get:49=2a-52a8a=18a-45a=92b2=454Substituting the values of a2 and b2, we get:x2a2+y2b2=14x281+4y245=120x2+36y2405=120x2+36y2=405This is the required equation of the ellipse.

(iii)  e=12 and semi major axis=4i.e. a=4We have e=1-b2a212=1-b216On squaring both sides, we get:14=16-b21616=64-4b2b2=12Substituting the values of a2 and b2, we get:x2a2+y2b2=1x216+y212=13x2+4y248=13x2+4y2=48This is the required equation of the ellipse.

(iv) e=12 and major axis=12i.e., 2a=12 or a=6We have e=1-b2a212=1-b236On squaring both sides, we get:14=36-b23636=144-4b2b2=27Substituting the values of a2 and b2, we get:x2a2+y2b2=1x236+y227=13x2+4y2108=13x2+4y2=108This is the required equation of the ellipse.

(v) The ellipse passes through 1,4 and -6,1.x2a2+y2b2=11a2+16b2=1Let 1a2=α and 1b2=βThen α+16β=1                   ..(1)It also passes through -6,1.x2a2+y2b2=136a2+1b2=136α+β=1       ...(2)Solving eqs. (1) and (2), we get:α=3115 and β=7115Substituting the values, we get: 3x2115+7y2115=13x2+7y2115=13x2+7y2=115This is the required equation of ellipse.
(vi) Vertices ±5,0 and focus ±4,0The coordinates of its vertices and foci are±a,0 and ±ae,0, respectively.i.e. a=5 and ae=4 e=45Now, b2=a21-e2b2=251-1625b2=9x225+y29=1This is the required equation of the ellipse.
(vii) Vertices 0,±13 and focus 0,±5The coordinates of its vertices and foci are0,±b and 0,±be, respectively.i.e. b=13 and be=5e=513Now, a2=b21-e2a2=1691-25169a2=144x2144+y2169=1This is the required equation of the ellipse.
(viii) Vertices ±6,0 and focus ±4,0The coordinates of its vertices and foci are ±a,0 and ±ae,0, respectively.i.e., a=6 and ae=4e=46or23Now, b2=a21-e2b2=361-1636b2=20 x236+y220=1This is the required equation of the ellipse.(ix) Let the equation of the ellipse be  x2a2+y2b2=1.  End of major axis=±3,0 End of minor axis=0,±2But the coordinates of the end points of the major and the minor axes are (±a,0) and 0,±b, respectively. a=3  and b=2Then x29+y24=1This is the required equation of the ellipse.(x)   Let the equation of the ellipse be  x2a2+y2b2=1.  End of major axis=0,±5 End of minor axis=±1,0But the coordinates of the end points of the major and the minor axes are (±a,0) and 0,±b, respectively.a=1  and b=5Then x21+y25=1This is the required equation of the ellipse.
(xi )Length of major axis=26Foci=±5,0We have 2a=26a=13Also, ae=5e=513Now, e=1-b2a2513=1-b2169On squaring both sides, we get:25169=169-b2169b2=144Now, x2a2+y2b2=1x2169+y2144=1This is the required equation of the ellipse.(xii) Length of minor axis=16 and foci=0,±6i.e. 2b=16b=8and be=6e=68Now, e=1-a2b268=1-a264On squaring both sides, we get: 3664=64-a264a2=28x264+y228=1This is the required equation of the ellipse.(xiii) Foci=±3,0 and a=4i.e. ae=3e=34Now, e=1-b2a234=1-b216On squaring both sides, we get: 916=16-b216b2=7x216+y27=1 This is the required equation of the ellipse.



Page No 26.23:

Question 6:

Find the equation of the ellipse whose foci are (4, 0) and (−4, 0), eccentricity = 1/3.

Answer:

Let the equation of the ellipse be x2a2+y2b2=1         ...(1)Then the coordinates of the foci are (±ae,0). Also, e=13We have ae=4a.13=4a=12Now, b2=a2(1-e2)b2=1221-132b2=14489b2=128Substituting the values of a2 and b2 in eq. (1), we get:x2144+y2128=1This is the required equation of the ellipse.

Page No 26.23:

Question 7:

Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

Answer:

According to the question, the minor axis is equal to the distance between the foci. i.e. 2b=2ae and 2b2a=10 or b2=5a b=aeb2=a2e2b2=a21-b2a2           e=1-b2a2b2=a2-b2a2=2b2a2=10a         b2=5a  a=10b2=5a b2=50Substituting the values of a and b in the equation of an ellipse, we get:x2100+y250=1x2+2y2=100This is the required equation of the ellipse.

Page No 26.23:

Question 8:

Find the equation of the ellipse whose centre is (−2, 3) and whose semi-axis are 3 and 2 when major axis is (i) parallel to x-axis (ii) parallel to y-axis.

Answer:

(i) When the major axis is parallel to the x-axis.Let (x-x1)2a2+(y-y1)2b2=1       ...(1)Here, xand y1 are -2 and 3, respectively, and 3 and 2 are the lengths of the axes. Substituting the value in eq. (1), we get: (x+2)29+(y-3)24=14(x2+4+4x)+9(y2+9-6y)36=14x2+16+16x+9y2+81-54y=364x2+9y2+16x-54y+61=0This is the required equation of the ellipse.(ii) When the major axis is parallel to the y-axis.Let (x-x1)2b2+(y-y1)2a2=1    ...(1)Here, x1 and y1 are -2 and 3, respectively, and 3 and 2 are the lengths of the axes.Substituting the value in eq. (1), we get:(x+2)24+(y-3)29=19(x2+4+4x)+4(y2+9-6y)36=19x2+36+36x+4y2+36-24y=369x2+4y2+36x-24y+36=0This is the required equation of the ellipse.

Page No 26.23:

Question 9:

Find the eccentricity of an ellipse whose latus rectum is
(i) half of its minor axis
(ii) half of its major axis.

Answer:

(i) According to the question, the latus rectum is half its minor axis. i.e.2b2a=12×2b2b2=ab2b=aNow, e=1-b24b2e=1-14e=322) According to the question, the latus rectum is half its minor axis. i.e. 2b2a=12×2a2b2=a2a2=2b2Now, e=1-b22b2e=1-12e=12

Page No 26.23:

Question 10:

Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:
(i) x2 + 2y2 − 2x + 12y + 10 = 0
(ii) x2 + 4y2 − 4x + 24y + 31 = 0
(iii) 4x2 + y2 − 8x + 2y + 1 = 0
(iv) 3x2 + 4y2 − 12x − 8y + 4 = 0
(v) 4x2 + 16y2 − 24x − 32y − 12 = 0
(vi) x2 + 4y2 − 2x = 0

Answer:

(i) x2+2y2-2x+12y+10=0x2-2x+2y2+6y=-10x2-2x+1+2y2+6y+9=-10+18+1x-12+2y+32=9x-129+y+3292=9Here, x1=1 and y1=-3Also, a=3 and b=32Centre= 1,-3Major axis=2a2×3=6Minor axis=2b2×32=32e=1-b2a2e=1-929e=12Foci= x1±ae,y1         =1±32,-3
(ii) x2+4y2-4x+24y+31=0x2-4x+4y2+6y=-31x2-4x+4+4y2+6y+9=-31+36+4x-22+4y+32=9x-229+y+3294=9Here, x1=2 and y1=-3 Also, a=3 and b=32Centre=x1,y1=2,-3Major axis=2a                 = 2×3                 =6Minor axis=2b                 =2×32                 =3e=1-b2a2e=1-949e=32Foci= x1±ae,y1       =2±332,-3
(iii) 4x2+y2-8x+2y+1=04x2-2x+y2+2y=-14x2-2x+1+y2+2y+1=-1+4+14x-12+y+12=4x-121+y+124=1Here, x1=1 and y1=-1 Also, a=1 and b=2Centre=x1,y1=1,-1Major axis=2b                =2×2                =4Minor axis=2a                =2×1                =2e=1-a2b2e=1-14e=32Foci =x1,y1±be         =1,-1±3
(iv)  3x2+4y2-12x-8y+4=03x2-4x+4y2-2y=-43x2-4x+4+4y2-2y+1=-4+12+43x-22+4y-12=12x-224+y-123=1Here, x1=2, y1=1Also, a=2 and b=3Centre=x1, y1=2,1Major axis=2a                =2×2                =4Minor axis=2b                =2×3                =23e=1-b2a2e=1-34e=12Foci= x1±ae,y1        =2±1,1
(v) 4x2+16y2-24x-32y-12=04x2-6x+16y2-2y=124x2-6x+9+16y2-2y+1=12+36+164x-32+16y-12=64x-3216+y-124=9Centre= 3,1Major axis=2a                  =2×4                  =8Minor axis=2b                  =2×2                  =4e=1-b2a2e=1-416e=32Foci= x±ae,y        =3±23,1(vi)  x2-2x+4y2=0x2-2x+4y2=0x2-2x+1+4y2=1x-12+4y2=1x-121+y214=9Centre= 1,0Major axis=2a                  =2×1                  =2Minor axis=2b                  =2×12                  =1e=1-b2a2e=1-141e=32Foci =x±ae,y         =1±32,0

Page No 26.23:

Question 11:

Find the equation of an ellipse whose foci are at (± 3, 0) and which passes through (4, 1).

Answer:

Let the equation of the ellipse be x2a2+y2b2=1         ... (1)Then ae=3 Also x=4 and y=1          [Ellipse passing through (4,1)]Substituting the values of x and y in eq. (1), we get: 42a2+12b2=116a2+1b2=1Now, b2=a2(1-e2)b2=a2-a2e2b2=a2-9 or a2=b2+9        ...216a2+1b2=116b2+a2=a2b216b2+b2+9=b2b2+917b2+9=b4+9b2b4-8b2-9=0b2-9b2+1b=±3Substituting the value of b in eq. (2), we get:a=32x218+y29=1This is the required equation of the ellipse.

Page No 26.23:

Question 12:

Find the equation of an ellipse whose eccentricity is 2/3, the latus-rectum is 5 and the centre is at the origin.

Answer:

Let the ellipse be x2a2+y2b2=1.       ...1e=23 and latus rectum=5     (Given)Now,2b2a=52b2=5a         ...(2)2a2(1-e2)=5a                  [b2=a2(1-e2)]2a21-49=5a2a2 ×59=5a10a2=45aa=92Substituting the value of a in eq. (2), we get:2b2=5×92b2=454Substituting the values of a2and b2in eq. (1), we get:x2814+y2454=14x281+4y245=1This is the required equation of the ellipse.

Page No 26.23:

Question 13:

Find the equation of an ellipse with its foci on y-axis, eccentricity 3/4, centre at the origin and passing through (6, 4).

Answer:

Let the equation of the ellipse bex2a2+y2b2=1 and let e=34. Also, let the foci be on the  y-axis.a2=b21-e2a2=b21-916a2=716b2x2a2+7y216a2=1      ...(1)It passes through 6,4.Then 36a2+11216a2=1576+11216a2=116a2=688a2=43Now, b2=16a27b2=6887Substituting the values of a2 and b2 in eq. (1), we get:x243+7y2688=1This is the required equation of the ellipse.

Page No 26.23:

Question 14:

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4, 3) and (−1, 4).

Answer:

Let the ellipse be x2a2+y2b2=1 and it passes through the points (4,3) and (-1,4).16a2+9b2=1and 1a2+16b2=1Let α=1a2 and β=1b2Then 16α+9β=1 and α+16β=1Solving these two equations, we get:α=7247 and β=152471a2=7247and 1b2=15247    ...(1) Substituting eq. (1) in the equation of an ellipse, we get:7x2247+15y2247=1This is the required equation of the ellipse.

Page No 26.23:

Question 15:

Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point (−3, 1) and has eccentricity equal to 2/5.

Answer:

Let the equation of the ellipse be x2a2+y2b2=1      ...(1) It passes through the point -3,1.∴ 9a2+1b2=1        ...( 2)and e=25Now, b2=a21-e2b2=a21-25b2=a2×35or3a25Substituting the value of b2 in eq. (2), we get:  9a2+53a2=127+53a2=1a2=323b2=3×3235 or 325Substituting the values of a and b in eq. (1), we get: 3x232+5y232=13x2+5y2=32This is the required equation of the ellipse.

Page No 26.23:

Question 16:

Find the equation of an ellipse, the distance between the foci is 8 units and the distance between the directrices is 18 units.

Answer:

The distance between the foci is 8 units.i.e. 2ae=8     ... (1)The distance between the directrices is 18 units.i.e. 2ae=18     ... (2)Comparing eqs. (1) and (2), we get:e=82aSubstituting the value of e in eq. (2), we get: 2a82a=184a2=18×8a2=36 or a =6Now, 2ae=812e=8 or e=23Then b2=a2(1-e2)b2=361-49b2=36×59b2=20x236+y220=1This is the required equation of ellipse.

Page No 26.23:

Question 17:

Find the equation of an ellipse whose vertices are (0, ± 10) and eccentricity e = 45.

Answer:

Let the equation of the required ellipse bex2a2+y2b2=1         ...(1)Since the vertices of the ellipse are on the y-axis, the coordinates of the vertices are (0,±b). b=10Now, a2=b2(1-e2)a2=1001-452a2=100×925a2=36Substituting the values of a2 and b2 in equation (1), we get: x236+y2100=1100x2+36y23600=1100x2+36y2=3600This is the required equation of the ellipse.

Page No 26.23:

Question 18:

A rod of length 12 m moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with x-axis.

Answer:

Let AB be the rod making an angle θ with OX and let P (x, y) be the point on it such that AP = 3 cm.
Then, PB = AB – AP = (12 – 3) cm = 9 cm      [∵ AB = 12 cm]
From P, draw PQ⊥OY and PR⊥OX.

In PBQ, we have:cos θ =PQPB=x9In PRA, we have:sin θ =PRPA=y3Since sin2 θ +cos2 θ=1, we have:y32+x92=1x281+y29=1Thus, the locus of a point P on the rod is x281+y29=1.

Page No 26.23:

Question 19:

Find the equation of the set of all points whose distances from (0, 4) are 23 of their distances from the line y = 9.

Answer:


We have
PQ=23PLx-02+y-42=23y-932x2+y-42=22y-929x2+9y2-72y+144=4y2-72y+3249x2+5y2=180x220+y236=1



Page No 26.27:

Question 1:

If the lengths of semi-major and semi-minor axes of an ellipse are 2 and 3 and their corresponding equations are y − 5 = 0 and x + 3 = 0, then write the equation of the ellipse.

Answer:

The lengths of the semi major and the semi minor axes of the ellipse are 2 and 3, respectively. Their corresponding equations are y-5=0 and x+3 =0.So, a=2 and b=3The equation of the ellipse is given byx+324+y-523=13x+32+4y-52=123x2+9+6x+4y2+25-10y=123x2+27+18x+4y2+100-40y=123x2+4y2+18x-40y+115=0

Page No 26.27:

Question 2:

Write the eccentricity of the ellipse 9x2 + 5y2 − 18x − 2y − 16 = 0.

Answer:

9x2+5y2-18x-2y-16=09x2-2x+5y2-2y5=169x2-2x+1+5y2-2y5+125=16+9+159x-12+5y-152=1265x-1212645+y-15212625=1a2=12645 and b2=12625Clearly, a<bNow, e=1-a2b2e=1-1264512625e=1-59e=23

Page No 26.27:

Question 3:

Write the centre and eccentricity of the ellipse 3x2 + 4y2 − 6x + 8y − 5 = 0.

Answer:

3x2-6x+4y2+8y-5=03(x2-2x)+4(y2+2y)=53(x2-2x+1)+4(y2+2y+1)=5+3+43(x-1)2+4(y+1)2=123(x-1)212+4(y+1)212=1(x-1)24+(y+1)23=1Compairing it withx2a2+y2b2=1, we get:a=2 and b=3Here, a>b, so the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively.Now, e=1-b2a2e=1-34e=14 e=12 and centre = 1,-1

Page No 26.27:

Question 4:

PSQ is a focal chord of the ellipse 4x2 + 9y2 = 36 such that SP = 4. If S' is the another focus, write the value of S'Q.

Answer:


The given equation of the ellipse is 4x2+9y2=36               ...(i)x29+y24=1This is of the form x2a2+y2b2=1, where a2=9 and b2=4 i.e a=3 and b=2Clearly a>b, therefore the major axis and minor axis of the ellipse are along x axis and y axis respectively .Let , e be the ecentricity of the ellipse . Then,  e=1-b2a2e=1-49        =53Therefore, coordinates of focus at S i.e ae, 0=5,0& coordinates of focus at S' =-ae, 0=-5,0It is given that PSQ is a focal chord As we know that , SP+S'P=2a4+S'P=6         SP=4S'P=2Let coordinates of P be m,nAs S'P=2n-02+m+52=2n2+m2+5+25m=4            ...(ii)& SP=4n-02+m-52=4n2+m2+5-25m=16     ...(iii)Subtracting eq (iii) from eq (ii) , we get 45m=-12m=-35& we get n=45 Coordinates of P are -35,45 and coordinates of S are 5,0 Equation of the line segment PS which is extended to PQ is given by x+2y=5             ...(iv)Solving eq (i) & (iv) we get , x=-35 and y=45 which are the coordinates of P& x= 66550 and y=-8550 which would be the coordinates of Q. S'Q=-8550-02+66550+52             =265 

Page No 26.27:

Question 5:

Write the eccentricity of an ellipse whose latus-rectum is one half of the minor axis.

Answer:

According to the question, the latus rectum is half its minor axis.i.e. 2b2a=12×2b2b=aNow, e=1-b2a2e=1-b24b2e=1-14e=32 

Page No 26.27:

Question 6:

If the distance between the foci of an ellipse is equal to the length of the latus-rectum, write the eccentricity of the ellipse.

Answer:

According to the question, the distance between the foci of an ellipse is equal to the length of the latus rectum. i.e. 2b2a=2aee=b2a2But e=1-b2a2Then e=1-e      Squaring both sides, we get:e2+e-1=0e=-1±1+42         Ecentricity cannot be negative e=5-12

Page No 26.27:

Question 7:

If S and S' are two foci of the ellipse x2a2+y2b2=1 and B is an end of the minor axis such that ∆BSS' is equilateral, then write the eccentricity of the ellipse.

Answer:


We know that the focal distance of a point B(0, b) is a±e.0 =ai.e. SB=SB'=aSB+SB'=2a Since BSS' is equilateral, we have:SB=SS'=S'B=2ae2ae+2ae=2a4ae=2ae=24e=12

Page No 26.27:

Question 8:

If the minor axis of an ellipse subtends an equilateral triangle with vertex at one end of major axis, then write the eccentricity of the ellipse.

Answer:



According to the question, the minor axis of the ellipse subtends an equilateral triangle with the vertex at one end of the major axis. AB=a2+b2 We know that ABC is an equilateral triangle. AB=BB'                                                                                a2+b2=2bOn squaring both sides, we have:a2+b2=4b2a2=3b2Now, e=1-b2a2e=1-b23b2e=1-13e=23

Page No 26.27:

Question 9:

If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse.

Answer:



According to the Pythogoras theorem, we have:OA2+OB2=AB2From the figure, we can see that OA=b2a-02+ae-02=b4a2+a2e2=OB and AB=2b2aNow, 2a2e2+b4a2=4b4a2a2e2+b4a2=2b4a2a2e2=-b4a2+2b4a2a2e2=b4a2e2=b4a4e=b2a2We know that e=1-b2a2e=1-eOn squaring both sides, we get: e2+e-1=0e=-1±1+42        Ecentricity cannot be negativee=5-12        

Page No 26.27:

Question 1:

For the ellipse 12x2 + 4y2 + 24x − 16y + 25 = 0

(a) centre is (−1, 2)

(b) lengths of the axes are 3 and 1

(c) eccentricity = 23

(d) all of these

Answer:

Disclaimer: The equation should be 12x2+4y2+24x-16y+24=0 instead of 12x2+4y2+24x-16y+25=0.(d) all of these12x2+4y2+24x-16y+24=012x2+2x+4y2-4y=-2412x2+2x+1+4y2-4y+4=-24+12+1612x+12+4y-22=4x+123+y-221=1So, the centre is -1,2.Here, a=3  and b=1The lengths of the axes are 3 and 1.Now, e=1-b2a2e=1-13e=23



Page No 26.28:

Question 2:

The equation of the ellipse with focus (−1, 1), directrix xy + 3 = 0 and eccentricity 1/2 is
(a) 7x2 + 2xy + 7y2 + 10x + 10y + 7 = 0
(b) 7x2 + 2xy + 7y2 + 10x − 10y + 7 = 0
(c) 7x2 + 2xy + 7y2 + 10x − 10y − 7 = 0
(d) none of these

Answer:

(b) 7x2+7y2+2xy-10y+10x+7=0

    

Let P(x,y) be any point on the ellipse whose focus and eccentricity are S-1,1 and e=12, respectively.Let PM be the perpendicular from P on the directrix. Then SP=e×PMSP=12×PM2SP=PM4SP2=PM24x+12+y-12=x-y+312+-1224x2+1+2x+y2+1-2y=x2+y2+9-2xy-6y+6x28x2+8+16x+8y2+8-16y=x2+y2+9-2xy-6y+6x7x2+7y2+2xy-10y+10x+7=0This is the required equation of the ellipse.

Page No 26.28:

Question 3:

The equation of the circle drawn with the two foci of x2a2+y2b2=1 as the end-points of a diameter is
(a) x2+ y2 = a2 + b2
(b) x2+ y2 = a2
(c) x2 + y2 = 2a2
(d) x2 + y2 = a2b2

Answer:

(d) x2+y2=a2-b2We have r=aeLet the equation of the circle be x2+y2=r2.Now,  x2+y2=a2e2          r=ae x2+y2=a21-b2a2                                          e=1-b2a2x2+y2=a2-b2 The required equation of the circle is x2+y2=a2-b2.

Page No 26.28:

Question 4:

The eccentricity of the ellipse x2a2+y2b2=1 if its latus rectum is equal to one half of its minor axis, is
(a) 12

(b) 32

(c) 12

(d) none of these

Answer:

(b) e=32According to the question, the latus rectum is half its minor axis.i.e. 2b2a=12×2b2b2=aba=2bNow, e=1-b2a2e=1-b24b2e=1-14e=32 

Page No 26.28:

Question 5:

The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is
(a) 5-12

(b) 5+12

(c) 5-14

(d) none of these

Answer:

(a) e=5-12    According to the question, the distance between the foci is equal to the length of the latus rectum.2b2a=2aeb2=a2eNow, e=1-b2a2e=1-a2ea2e=1-eOn squaring both sides, we get:e2+e-1=0e=-1±1+42e=5-12       e cannot be negative

Page No 26.28:

Question 6:

The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is
(a) 32

(b) 23

(c) 12

(d) 23

Answer:

(c) 12According to the question, the minor axis is equal to the distance between the foci.i.e. 2b=2aee=ba       ...(1) Now, e=1-b2a2e=1-e2         [From (1)] On squaring both sides, we get:e2=1-e22e2=1e2=12e=12

Page No 26.28:

Question 7:

The difference between the lengths of the major axis and the latus-rectum of an ellipse is
(a) ae
(b) 2ae
(c) ae2
(d) 2ae2

Answer:

(d) 2ae2Length of the latus rectum=2b2aand e=1-b2a2a2e2=a2-b2b2=a2-a2e2b2=a21-e2 Length of the latus rectum=2a21-e2a=2a(1-e2)Length of the major axis=2aDifference between length of latus rectum and length of major axis = 2a-2a(1-e2)                                                                                                                          =2a-2a+2ae2                                                                                                                     =2ae2

Page No 26.28:

Question 8:

The eccentricity of the conic 9x2 + 25y2 = 225 is
(a) 2/5
(b) 4/5
(c) 1/3
(d) 1/5
(e) 3/5

Answer:

(b) 459x2+25y2=225x225+y29=1Comparing it with x2a2+y2b2=1, we get: a=5 and b=3Here, a>b, so the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively.Now, e=1-b2a2e=1-925e=1625e=45

Page No 26.28:

Question 9:

The latus-rectum of the conic 3x2 + 4y2 − 6x + 8y − 5 = 0 is

(a) 3

(b) 32

(c) 23

(d) none of these

Answer:

(a) 33x2+4y2-6x+8y-5=03(x2-2x)+4(y2+2y)=53(x2-2x+1)+4(y2+2y+1)=5+3+43(x-1)2+4(y+1)2=12(x-1)24+(y+1)23=1So, a=2 and b=3Latus rectum=2b2a                           =2322                           =3

Page No 26.28:

Question 10:

The equations of the tangents to the ellipse 9x2 + 16y2 = 144 from the point (2, 3) are
(a) y = 3, x = 5
(b) x = 2, y = 3
(c) x = 3, y = 2
(d) x + y = 5, y = 3

Answer:

(d) x+y=5 , y=39x2+16y2=144x216+y29=1Equation of the tangent in case of an ellipse is given by y=mx+a2m2+b2y=mx+16m2+9              ... (1) Substituting  x=2 and y=3, we get:3=2m±16m2+93-2m=16m2+9On squaring both sides, we get: 3-2m2=16m2+99+4m2-12m=16m2+912m2+12m=012mm+1=0m=0,-1Substituting values of m in eq. (1), we get: For m=0, y=3For m=-1, y=-x+5 or x+y=5                                       

Page No 26.28:

Question 11:

The eccentricity of the ellipse 4x2 + 9y2 + 8x + 36y + 4 = 0 is

(a) 56

(b) 35

(c) 23

(d) 53

Answer:

(d) 534x2+9y2+8x+36y+4=04(x2+2x)+9(y2+4y)=-44(x2+2x+1)+9(y2+4y+4)=-4+4+364(x+1)2+9(y+2)2=364(x+1)236+9(y+2)236=1(x+1)29+(y+2)24=1Comparing it with x2a2+y2b2=1, we get:a=3 and b=2So, the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively.Now, e=1-b2a2e=1-49e=59e=53

Page No 26.28:

Question 12:

The eccentricity of the ellipse 4x2 + 9y2 = 36 is

(a) 123

(b) 13

(c) 53

(d) 56

Answer:

(c) 534x2+9y2=36x29+y24=1Comparing it with x2a2+y2b2=1, we get:a=3 and b=2Here, a>b, so the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively.Now, e=1-b2a2e=1-49e=59e=53

Page No 26.28:

Question 13:

The eccentricity of the ellipse 5x2 + 9y2 = 1 is
(a) 2/3
(b) 3/4
(c) 4/5
(d) 1/2

Answer:

(a) 235x2+9y2=1x25+y29=1 a=5 and b=3Here, b>a, so the major and the minor axes of the ellipse are along the x-axis and the y-axis, respectively.Now, e=1-a2b2e=1-59e=49e=23

Page No 26.28:

Question 14:

For the ellipse x2 + 4y2 = 9
(a) the eccentricity is 1/2
(b) the latus-rectum is 3/2
(c) a focus is 33, 0
(d) a directrix is x = -23

Answer:

(b) the latus rectum is 32x2+4y2=9x29+y294=1Here,a>b e=1-b2a2e=1-994e=1-14e=32Latus rectum=2b2a                     =2×943                     =32Focus =±ae,0              =332,0Directrix x=±ae              =±332               =23



Page No 26.29:

Question 15:

If the latus rectum of an ellipse is one half of its minor axis, then its eccentricity is

(a) 12

(b) 12

(c) 32

(d) 34

Answer:

(c) 32According to the question, the latus rectum of an ellipse is half its minor axis.i.e. 2b2a=12×2b2b2=aba=2bNow, e=1-b2a2e=1-b24b2e=1-14e=34e=32

Page No 26.29:

Question 16:

An ellipse has its centre at (1, −1) and semi-major axis = 8 and it passes through the point
(1, 3). The equation of the ellipse is

(a) x+1264+y+1216=1

(b) x-1264+y+1216=1

(c) x-1216+y+1264=1

(d) x+1264+y-1216=1

Answer:

(b) x-1264+y+1216=1According to the question, the centre is at 1,-1.a=8      Givenx-12a2+y+12b2=1      ...(1)It passes through point 1,3.i.e. x=1 and y=3Putting these values in eq. (1), we get:1-12a2+3+12b2=116b2=1b2=16 or b=4Substituting the values of a and b in eq. (1), we get: x-1264+y+1216=1

Page No 26.29:

Question 17:

The sum of the focal distances of any point on the ellipse 9x2 + 16y2 = 144 is
(a) 32
(b) 18
(c) 16
(d) 8

Answer:

(d) 89x2+ 16y2= 144x216+y29=1Here, a=4 and b=3The sum of the focal distances of any point on an ellipse is constant and equal to the length of the major axis of the ellipse.i.e. sum of the focal distances of any point on an ellipse=2a                                                                                 =2×4                                                                                 =8

Page No 26.29:

Question 18:

If (2, 4) and (10, 10) are the ends of a latus-rectum of an ellipse with eccentricity 1/2, then the length of semi-major axis is
(a) 20/3
(b) 15/3
(c) 40/3
(d) none of these

Answer:

(a) 203e=1 2       GivenNow, e=1-b2a212=1-b2a2On squaring both sides, we get:14=1-b2a214=a2-b2a24a2-4b2=a23a2=4b2b2a2=34a2=4b23or a=2b3          ...(1)Latus rectum=2b2a2If (2,4) and (10,10) are the ends points of a latus rectum.       i.e. 10-22+10-42=2b2×32b64+36=3bOn squaring both sides, we get:100=3b2b=103Now, a=2b3a=203×13a=203So, the length of the semi major axis is 203.

Page No 26.29:

Question 19:

The equation x22-λ+y2λ-5+1=0 represents an ellipse, if
(a) λ < 5
(b) λ < 2
(c) 2 < λ < 5
(d) λ < 2 or λ > 5

Answer:

(c) 2 <λ <5x22-λ+y2λ-5+1=0x2λ-2+y25-λ=1To represent the equation of ellipse, we have: λ-2>0λ>2and5-λ>05<λ∴ 2<λ<5

Page No 26.29:

Question 20:

The eccentricity of the ellipse 9x2 + 25y2 − 18x − 100y − 116 = 0, is
(a) 25/16
(b) 4/5
(c) 16/25
(d) 5/4

Answer:

(b) 459x2-18x+25y2-100y-116=09(x2-2x)+25(y2-4y)=1169(x2-2x+1)+25(y2-4y+4)=116+100+99(x-1)2+25(y-2)2=2259(x-1)2225+25(y-2)2225=1(x-1)225+(y-2)29=1Comparing it with x2a2+y2b2=1, we get:a=5 and b=3Here, a>b, so the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively.Now, e=1-b2a2e=1-925e=1625e=45

Page No 26.29:

Question 21:

If the major axis of an ellipse is three times the minor axis, then its eccentricity is equal to
(a) 13

(b) 13

(c) 12

(d) 223

(e) 232

Answer:

(d) 223Length of the major axis =2bLength of the minor axis=2aAccording to question, the major axis of the ellipse is three times the minor axis.i.e. 2b=3(2a)2b=6aa=2b/6a=b/3, b=3aHere, a<b, so the major and the minor axes of the ellipse are along the x-axis and the y-axis, respectively.Now, e=1-a2b2e=1-b29b2e=1-19e=89e=223

Page No 26.29:

Question 22:

The eccentricity of the ellipse 25x2 + 16y2 = 400 is
(a) 3/5
(b) 1/3
(c) 2/5
(d) 1/5

Answer:

(a) 3525x2+16y2=400x216+y225=1     ...(1)Comparing equation (1)  with x2a2+y2b2=1, we get: a2=16 and b2=25Here, a<b, so the major and the minor axes of the given ellipse are along the y-axis and the x-axis, respectively.Now, e=1-a2b2e=1-1625e=925e=35

Page No 26.29:

Question 23:

The eccentricity of the ellipse 5x2 + 9y2 = 1 is
(a) 2/3
(b) 3/4
(c) 4/5
(d) 1/2

Answer:

(a) 235x2+9y2=1x215+y219=1Comparing with the standard equation of the ellipse, we get: a2=15 and b2=19, i.e. a=15 and b=13Here, a>b Now, e=1-b2a2e=1-1915e=1-59e=49e=23

Page No 26.29:

Question 24:

The eccentricity of the ellipse 4x2 + 9y2 = 36 is

(a) 123

(b) 13

(c) 53

(d) 56

Answer:

(c)  53
4x2+9y2=36x29+y24=1      .. (1)Compairing equation (1 ) with x2a2+y2b2=1, we geta2=9 and b2=4Here, a>b, so the major and the minor axes of the given ellipse are along the x-axis and the y-axis, respectively.Now, e=1-b2a2e=1-49e=59e=53



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