Question

Let P (a sec q, b tan q) and Q (a sec f, b tan f) where

q + f = p/2, be two points on the hyperbola x²/a² – y²/b² = 1. If (h, k) is the point of intersection of normals at P and Q, then k is equal to –

• A. $\frac{a^{2} + b^{2}}{a}$
• B. –$\left\lbrack \frac{a^{2} + b^{2}}{a} \right\rbrack$
• C. $\frac{a^{2} + b^{2}}{b}$
• D. –$\left\lbrack \frac{a^{2} + b^{2}}{b} \right\rbrack$

Answer 1

Answer: (D)

–$\left\lbrack \frac{a^{2} + b^{2}}{b} \right\rbrack$

Answer 2

Equation of the tangent at P (a sec q, b tan q) is

$\frac{x}{a}$sec q – $\frac{y}{a}$ tan q = 1.

Therefore equation of the normal at P is

y – b tan q = –$\frac{a}{b}$ sin q (x – a sec q)

Ž ax + b cosec q y = (a² + b²) sec q …(1)

Similarly the equation of the normal at

Q (a sec f, b sec f) is

ax + b cosec f y = (a² + b²) sec f … (2)

Subtracting (2) from (1) we get y = $\frac{a^{2} + b^{2}}{b}$.

$\frac{\sec\theta - \sec\varphi}{\cos ec\theta - \cos ec\varphi}$

So that k = y = $\frac{a^{2} + b^{2}}{b}$

$\frac{\sec\theta - \sec(\pi/2 - \theta)}{\cos ec\theta - \cos ec(\pi/2 - \theta)}$

[Q q + f = p/2]

= $\frac{a^{2} + b^{2}}{b}$ $\frac{\sec\theta - \cos ec\theta}{\cos ec\theta - \sec\theta}$ = – $\left\lbrack \frac{a^{2} + b^{2}}{b} \right\rbrack$.