Match the items of Column – I with Column – II Column – I Column – IIIA tangent drawn to hyperbola x2a2−y2b2=1 at P(π6) forms a triangle of area 3a2 square units, with coordinate axes, then the square of its eccentricity is equal to P17IIIf the eccentricity of the hyperbola x2−y2sec2θ=5 is 3 times the eccentricity of the ellipse x2sec2θ+y2=25 , then 4cos2θ is Q8IIIFor the hyperbola x23−y2=3 , acute angle between its asymptotes is Lπ24 , then value of ‘L’ is R16IVFor the hyperbola xy =8 any tangent of it at P meets co-ordinate axes at Q and R then area of triangle CQR where ‘C’ is centre of the hyperbola is S2 T5

I.The point P(π6) is (asecπ6,tanπ6) i.e. P(2a3,b3)∴ Equation of tangent at P is x3a2−y3b=1∴ area of the triangle 12×3a2×3b=3a2 ∴ ba=4 ∴ e2=1+b2a2=17 II.eccentricity of the hyperbola x2−y2sec2θ=5 is e1=1+sec2θsec2θ=1+cos2θEccentricity of the ellipse x2sec2θ+y2=25 is e2=sec2θ−1sec2θ=|sinθ|e1=3e2 ⇒1+cos2θ=3sin2θ ⇒cosθ=±12 ∴ least positive value of θ is π4 ∴ P=24III.Asymptotes are x=±3y∴ angle between the asymptotes is π3 ∴ l=8 IV. any point of xy = 8 is P(8t,8t)∴ equation of the tangent at P is x16t8+y168t=1∴ area of the triangle =12.16t8.68t=16

Match the items of Column – I with Column – II Column – I Column – IIIA tangent drawn to hyperbola x2a2−y2b2=1 at P(π6) forms a triangle of area 3a2 square units, with coordinate axes, then the square of its eccentricity is equal to P17IIIf the eccentricity of the hyperbola x2−y2sec2θ=5 is 3 times the eccentricity of the ellipse x2sec2θ+y2=25 , then 4cos2θ is Q8IIIFor the hyperbola x23−y2=3 , acute angle between its asymptotes is Lπ24 , then value of ‘L’ is R16IVFor the hyperbola xy =8 any tangent of it at P meets co-ordinate axes at Q and R then area of triangle CQR where ‘C’ is centre of the hyperbola is S2 T5

I.The point P(π6) is (asecπ6,tanπ6) i.e. P(2a3,b3)∴ Equation of tangent at P is x3a2−y3b=1∴ area of the triangle 12×3a2×3b=3a2 ∴ ba=4 ∴ e2=1+b2a2=17 II.eccentricity of the hyperbola x2−y2sec2θ=5 is e1=1+sec2θsec2θ=1+cos2θEccentricity of the ellipse x2sec2θ+y2=25 is e2=sec2θ−1sec2θ=|sinθ|e1=3e2 ⇒1+cos2θ=3sin2θ ⇒cosθ=±12 ∴ least positive value of θ is π4 ∴ P=24III.Asymptotes are x=±3y∴ angle between the asymptotes is π3 ∴ l=8 IV. any point of xy = 8 is P(8t,8t)∴ equation of the tangent at P is x16t8+y168t=1∴ area of the triangle =12.16t8.68t=16

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Match the items of Column – I with Column – II
Column – I Column – II
I A tangent drawn to hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at $P (\frac{π}{6})$ forms a triangle of area $3 a^{2}$ square units, with coordinate axes, then the square of its eccentricity is equal to P 17
II If the eccentricity of the hyperbola $x^{2} - y^{2} \sec^{2} θ = 5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^{2} \sec^{2} θ + y^{2} = 25$ , then $4 \cos^{2} θ$ is Q 8
III For the hyperbola $\frac{x^{2}}{3} - y^{2} = 3$ , acute angle between its asymptotes is $\frac{L π}{24}$ , then value of ‘L’ is R 16
IV For the hyperbola $x y = 8$ any tangent of it at P meets co-ordinate axes at Q and R then area of triangle CQR where ‘C’ is centre of the hyperbola is S 2
T 5

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Detailed Solution

I.The point $P (\frac{π}{6})$ is $(a \sec \frac{π}{6}, \tan \frac{π}{6})$ i.e. $P (\frac{2 a}{\sqrt{3}}, \frac{b}{\sqrt{3}})$
$∴$ Equation of tangent at P is $\frac{x}{\frac{\sqrt{3} a}{2}} - \frac{y}{\sqrt{3} b} = 1$
$∴$ area of the triangle $\frac{1}{2} \times \frac{\sqrt{3} a}{2} \times \sqrt{3} b = 3 a^{2}$
$∴$ $\frac{b}{a} = 4$
$∴$    $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 17$
II.eccentricity of the hyperbola $x^{2} - y^{2} \sec^{2} θ = 5$ is $e_{1} = \sqrt{\frac{1 + \sec^{2} θ}{\sec^{2} θ}} = \sqrt{1 + \cos^{2} θ}$
Eccentricity of the ellipse $x^{2} \sec^{2} θ + y^{2} = 25$ is $e_{2} = \sqrt{\frac{\sec^{2} θ - 1}{\sec^{2} θ}} = | \sin θ |$
$e_{1} = \sqrt{3} e_{2} \Rightarrow 1 + \cos^{2} θ = 3 \sin^{2} θ \Rightarrow \cos θ = \pm \frac{1}{\sqrt{2}}$
$∴$ least positive value of $θ$ is $\frac{π}{4}$ $∴$ P=24
III.Asymptotes are $x = \pm \sqrt{3} y$
$∴$ angle between the asymptotes is $\frac{π}{3}$    $∴$ $l = 8$
IV. any point of xy = 8 is $P (\sqrt{8} t, \frac{\sqrt{8}}{t})$
$∴$ equation of the tangent at P is $\frac{x}{\frac{16 t}{\sqrt{8}}} + \frac{y}{\frac{16}{\sqrt{8} t}} = 1$
$∴$ area of the triangle $= \frac{1}{2} . \frac{16 t}{\sqrt{8}} . \frac{6}{\sqrt{8} t} = 16$

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