Question

C the centre of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$ = 1. The tangents at any point P on this hyperbola meets the straight lines bx – ay = 0 and bx + ay = 0 in the points Q and R respectively. Then CQ. CR =

• A. a² + b²
• B. a² – b²
• C. $\frac{1}{a^{2}} + \frac{1}{b^{2}}$
• D. $\frac{1}{a^{2}} - \frac{1}{b^{2}}$

Answer 1

Answer: (A)

a² + b²

Answer 2

P is (a sec θ, b tan θ)

Tangent t at P is $\frac{x\sec\theta}{a} - \frac{y\tan\theta}{b}$ = 1

It meets bx – ay = 0 i.e., $\frac{x}{a} = \frac{y}{b}$in Q

∴ Q is $\left( \frac{a}{\sec\theta - \tan\theta},\frac{- b}{\sec\theta - \tan\theta} \right)$

It meets bx + ay = 0 i.e., $\frac{x}{a} = - \frac{y}{b}$ in R.

∴ R is $\left( \frac{a}{\sec\theta + \tan\theta},\frac{- b}{\sec\theta + \tan\theta} \right)$

∴ CQ .CR = $\frac{\sqrt{a^{2} + b^{2}}}{\left( \sec\theta - \tan\theta \right)}.\frac{\sqrt{a^{2} + b^{2}}}{\left( \sec\theta + \tan\theta \right)}$

= a² + b², {$\because$ sec² θ tan² θ= 1}