Question

The tangent at P on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets one of the asymptote in Q. If the locus of the mid points of PQ is a hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\lambda ,$ find the value of $4\lambda$.

Answer 1

Given hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$-------------(1)

Let there be a point P$(a\sec \theta ,b\tan \theta )$ on hyperbola

Tangent from P:$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$---------------(2)

Meets one asymplote in Q.

Locus of mid point of PQ is another hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\lambda$-------------(3)

Let the midpoint be M

Point of intersectionof equation(2) and $y=\frac{b}{a}x$ is Q:$\Rightarrow \frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$ $\Rightarrow x=\frac{a}{\sec \theta -\tan \theta }$

and now $y=\frac{b}{\sec \theta -\tan \theta }$ $\Rightarrow Q:(\frac{a}{\sec \theta -\tan \theta },\frac{b}{\sec \theta -\tan \theta })$

Now midpoint of PQ is M : $(\frac{\frac{a}{\sec \theta -\tan \theta }+a\sec \theta }{2},\frac{b}{\sec \theta -\tan \theta }+b\tan \theta )$

M : $(a\sec \theta +a\tan \theta +a\sec \theta ,b\sec \theta +b\tan \theta +b\tan \theta )$

$(as\sec \theta +\tan \theta =\frac{1}{\sec \theta -\tan \theta })$

Let the mid point of PQ be (h,k)

M lies on the equation (3) $\Rightarrow \frac{{(2a\sec \theta +a\tan \theta )}^2}{a^2}-\frac{{(b\sec \theta +2b\tan \theta )}^2}{b^2}=\lambda$

$\Rightarrow \lambda =\frac{4a^2{\sec }^2\theta +a^2{\tan }^2\theta +4a^2\sec \theta \tan \theta }{a^2}-\frac{b^2{\sec }^2\theta +4b^2{\tan }^2\theta +4b^2\sec \theta \tan \theta }{b^2}$

$\Rightarrow \lambda =4{\sec }^2\theta +{\tan }^2\theta +4\sec \theta \tan \theta -{\sec }^2\theta -4{\tan }^2\theta -4\sec \theta \tan \theta$

$\Rightarrow \lambda =4+(-1)$

$\Rightarrow \lambda =3$

Value of $4\lambda =12$