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)

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

Consider the equation of the parabola.

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

We know that the equation of the tangent from any point $M(a\sec \theta ,b\sin \theta )$ to the hyperbola is,

$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$

Therefore,

Coordinates of $A=(0,\frac{-b}{\tan \theta })$

Coordinates of $B=(\frac{a}{\sec \theta },0)$

Now, consider rectangle $OAPB$. Let the point be $P(h,k)$. Then,

$AP=OB$

$\Rightarrow h=\frac{a}{\sec \theta }$

$\sec \theta =\frac{a}{h}$

$PB=OA$

$\Rightarrow k=\frac{-b}{\tan \theta }$

$\tan \theta =\frac{-b}{k}$

We know that,

${\sec }^2\theta -{\tan }^2\theta =1$

$\frac{a^2}{h^2}-\frac{b^2}{k^2}=1$

Now, substitute $(x,y)$ for $(h,k)$.

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

Hence, this is the required locus of the point.