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Question: From any point on the hyperbola \(\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}}\) = 1, tangents are drawn ...

From any point on the hyperbola x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 1, tangents are drawn to the hyperbola x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 2. The area cut off by the chord of contact on the region between the asymptotes is equal to-

A

ab2\frac{ab}{2}

B

ab

C

2ab

D

4ab

Answer

4ab

Explanation

Solution

Let P (a sec q, b tan q) be any point on the hyperbola x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 1. Equation of the chord of contact of tangents from P to the hyperbola

x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 2 is

xasec⁡θa2–ybtan⁡θb2\frac{xa\sec\theta}{a^{2}}–\frac{yb\tan\theta}{b^{2}} = 2

or xsec⁡θa–ytan⁡θb\frac{x\sec\theta}{a}–\frac{y\tan\theta}{b} = 2 ......(1)

The two hyperbolas have a common set of asymptotes

y = ± ba\frac{b}{a}x.

y = ba\frac{b}{a}x meets the chord of contact of tangents at Q (2asec⁡θ–tan⁡θ,2bsec⁡θ–tan⁡θ)\left( \frac{2a}{\sec\theta –\tan\theta},\frac{2b}{\sec\theta –\tan\theta} \right)

y = – ba\frac{b}{a}x meets the chord of contact at

R(2asec⁡θ+tan⁡θ,–2bsec⁡θ+tan⁡θ)\left( \frac{2a}{\sec\theta + \tan\theta},\frac{–2b}{\sec\theta + \tan\theta} \right)

Area of DOQR

= 12\frac { 1 } { 2 } ∣0012asec⁡θ–tan⁡θ2bsec⁡θ–tan⁡θ12asec⁡θ+tan⁡θ–2bsec⁡θ+tan⁡θ1∣\left| \begin{matrix} 0 & 0 & 1 \\ \frac{2a}{\sec\theta –\tan\theta} & \frac{2b}{\sec\theta –\tan\theta} & 1 \\ \frac{2a}{\sec\theta + \tan\theta} & \frac{–2b}{\sec\theta + \tan\theta} & 1 \end{matrix} \right|

= 4ab sq. units