Question

If one axis of varying standard hyperbola be fixed in magnitude and position, then the locus of the point of contact of tangents drawn to it from a fixed point $(0,c)$ on the other axis is

• A. $y^2=\frac{a^2}{c}(x-c)$
  
• B. $x^2=-\frac{a^2}{c}(y-c)$
• C. $x^2=\frac{a^2}{c}(y+c)$
  
• D. $y^2=-\frac{a^2}{c}(x-c)$
  

Answer 1

The correct option is B $x^2=-\frac{a^2}{c}(y-c)$

Let transverse axis of hyberbola whose length is $2a$ and a point on conjugate axis $(0,c)$ be fixed.
Now ,Equation of a tangent at $P\left(\theta \right)$ is
$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\cdots \left(i\right)$
Let locus of required point be $h=a\sec \theta ,k=b\tan \theta$
$\because$ equation $\left(i\right)$ passes through the point $(0,c)$
$\Rightarrow -c\tan \theta =b\cdots \left(ii\right)$



Now, substituting the value of $b$ in $h=a\sec \theta ,k=b\tan \theta$ and then eliminating $\theta ,b$ we obtain required locus equation
$\frac{k}{c}=-{\tan }^2\theta \cdots \left(iii\right)$
and $\frac{h^2}{a^2}={\sec }^2\theta \cdots \left(iv\right)$
On adding Eqs. $\left(iii\right)$ and $\left(iv\right)$, $\frac{k}{c}+\frac{h^2}{a^2}=1\Rightarrow \frac{h^2}{a^2}=1-\frac{k}{c}$
$\Rightarrow h^2=\frac{a^2}{c}(c-k)$
$\therefore x^2=-\frac{a^2}{c}(y-c)$