Question

Tangent at any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ cut the axes at $A$ and $B$ respectively. If the rectangle OAPB (where $O$ is origin) is completed then locus of point $P$ is given by

• A. $\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$
• B. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$
• C. $\frac{a^2}{y^2}-\frac{b^2}{x^2}=1$
• D. none of these

Answer 1

Answer: (A)

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
tangent at point $P$ ( $a \sec \theta, b \tan \theta$ )
$\frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1 \text { or } \frac{x}{a \cos \theta}+\frac{y}{(-b \cot \theta)}=1$
Point A(acos $\theta, 0), B(0,-b \cot \theta)$
Cordinate of point $P$ is
$(h, k) \equiv(a \cos \theta,-b \cot \theta)$
$\cos \theta=\frac{h}{a}, \cot \theta=-\frac{k}{b} $
$ \cot \theta=\frac{h}{\sqrt{a^2-h^2}}=-\frac{k}{b}$
$ \frac{h^2}{a^2-h^2}=\frac{k^2}{b^2} $
$\frac{a^2}{h^2}-1=\frac{b^2}{k^2}$
So locus is
$\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$