Let C be the centre of the hyperbola frac{x^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the point $P$ on the hyperbola be $(a \sec 	heta, b 	an 	heta)$.The equation of the tangent at $P$ is $\frac{x \sec 	heta}{a} - \frac{y

Vedclass · Accepted Answer

$a^2+b^2$

Vedclass · Answer

(B) Let the point $P$ on the hyperbola be $(a \sec 	heta, b 	an 	heta)$.The equation of the tangent at $P$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$.The lines are $L_1: bx - ay = 0$ and $L_2: bx + ay = 0$.To find $Q$,solve $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$ and $bx = ay \implies y = \frac{bx}{a}$.Substituting $y$: $x(\frac{\sec 	heta}{a} - \frac{	an 	heta}{a}) = 1 \implies x = \frac{a}{\sec 	heta - 	an 	heta} = a(\sec 	heta + 	an 	heta)$.Then $y = b(\sec 	heta + 	an 	heta)$. So $Q = (a(\sec 	heta + 	an 	heta), b(\sec 	heta + 	an 	heta))$.$CQ^2 = a^2(\sec 	heta + 	an 	heta)^2 + b^2(\sec 	heta + 	an 	heta)^2 = (a^2+b^2)(\sec 	heta + 	an 	heta)^2$.Similarly,for $R$,solve with $bx = -ay \implies y = -\frac{bx}{a}$.$x(\frac{\sec 	heta}{a} + \frac{	an 	heta}{a}) = 1 \implies x = \frac{a}{\sec 	heta + 	an 	heta} = a(\sec 	heta - 	an 	heta)$.Then $y = -b(\sec 	heta - 	an 	heta)$. So $R = (a(\sec 	heta - 	an 	heta), -b(\sec 	heta - 	an 	heta))$.$CR^2 = (a^2+b^2)(\sec 	heta - 	an 	heta)^2$.$CQ \cdot CR = \sqrt{(a^2+b^2)(\sec 	heta + 	an 	heta)^2} \cdot \sqrt{(a^2+b^2)(\sec 	heta - 	an 	heta)^2} = (a^2+b^2)|\sec^2 	heta - 	an^2 	heta| = a^2+b^2$.

Let $C$ be the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $P$ be a point on it. If the tangent at $P$ to the hyperbola meets the straight lines $bx-ay=0$ and $bx+ay=0$ respectively in $Q$ and $R$,then $CQ \cdot CR=$

Similar Questions

Find the eccentricity of the hyperbola $16x^{2} - 32x - 3y^{2} + 12y = 44$.

The tangent to the hyperbola $x^2 - 3y^2 = 3$ at the point $(\sqrt{3}, 0)$ when associated with its two asymptotes constitutes:

If the length of the transverse and conjugate axes of a hyperbola are $8$ and $6$ respectively,then the difference of the focal distances of any point on the hyperbola will be

Let $H$ be the hyperbola,whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $e = \sqrt{2}$. Then the length of its latus rectum is

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $e = \frac{2}{\sqrt{7}-\sqrt{3}}$,then $\sin \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module