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
$\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$