Conic Sections
Let $H:\frac{{x}^{2}}{{a}^{2}}\frac{{y}^{2}}{{b}^{2}}=1,$ where a > b > 0, be a hyperbola in the xyplane whose conjugate axis LM subtends an angle of 60° at one of its vertices N. Let the area of the triangle LMN be $4\sqrt{3}$.
The correct option is:
LIST–I  LIST–II  
P.  The length of the conjugate axis of H is  1.  8 
Q.  The eccentricity of H is  2.  $\frac{4}{\sqrt{3}}$ 
R  The distance between the foci of H is  3.  $\frac{2}{\sqrt{3}}$ 
S.  The length of the latus rectum of H is  4.  4 
The correct option is:
View Solution
JEE Advanced 2018
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$  (Q) $\left(\frac{\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ 
(III) y^{2} = 4ax  (iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}$  (R) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$  (S) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}}\right)$ 
If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8,16), then which of the following options is the only CORRECT combination ?
View Solution
JEE Advanced 2017
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$  (Q) $\left(\frac{\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ 
(III) y^{2} = 4ax  (iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}$  (R) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$  (S) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}}\right)$ 
For $\mathrm{a}=\sqrt{2}$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact (–1,1), then which of the following options is the only CORRECT combination for obtaining its equation ?
View Solution
JEE Advanced 2017
Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively.  
Column1  Column2  Column3 
(I) x^{2} + y^{2} = a^{2}  (i) my = m^{2}x + a  (P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ 
(II) x^{2} + a^{2}y^{2} = a^{2}  (ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$  (Q) $\left(\frac{\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ 
(III) y^{2} = 4ax  (iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}$  (R) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ 
(IV) x^{2} – a^{2}y^{2} = a^{2}  (iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$  (S) $\left(\frac{{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}1}}\right)$ 
The tangent to a suitable conic (Column 1) at $\left(\sqrt{3},\frac{1}{2}\right)$ is found to be $\sqrt{3}\mathrm{x}+2\mathrm{y}=4$, then which of the following options is the only CORRECT combination?
View Solution
JEE Advanced 2017
PARAGRAPH 2
Let F_{1}(x_{1}, 0) and F_{2}(x_{2}, 0), for x_{1} < 0 and x_{2} > 0, be the foci of the ellips $\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{8}=1.$
Suppose a parabola having vertex at the origin and focus at F_{2} intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. 
If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the xaxis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF_{1}NF_{2} is
View Solution
JEE Advanced 2016
PARAGRAPH 2
Let F_{1}(x_{1}, 0) and F_{2}(x_{2}, 0), for x_{1} < 0 and x_{2} > 0, be the foci of the ellips $\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{8}=1.$
Suppose a parabola having vertex at the origin and focus at F_{2} intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. 
The orthocentre of the triangle F_{1}MN is
View Solution
JEE Advanced 2016
View Solution
JEE Advanced 2012
View Solution
JEE 2010

Column I 

Column II 
(A) 
Circle 
(p) 
The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x^{2} + y^{2} = 4 
(B) 
Parabola 
(q) 
Points z in the complex plane satisfying 
(C) 
Ellipse 
(r) 
Points of the conic have parametric representation 
(D) 
Hyperbola 
(s) 
The eccentricity of the conic lies in the interval 


(t) 
Points z in the complex plane satisfying 
View Solution
JEE 2009
View Solution
JEE 2009
View Solution
JEE 2009
View Solution
JEE 2009
Match the statements in Column I with the properties in Column II and tick the correct answer in the 4 × 4 matrix given in the ORS.
Column I  Column II  
(A)  Two intersecting circles  (p)  have a common tangent 
(B)  Two mutually external circles  (q)  have a common normal 
(C)  Two circles, one strictly inside the other  (r)  do not have a common tangent 
(D)  Two branches of a hyperbola  (s)  do not have a common normal 
View Solution
JEE 2007