Question

If the locus of midpoints of portions of tangents intercepted between co-ordinate axes of hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ is $\frac{\alpha }{x^2}-\frac{9}{y^2}=\beta$, then $(\alpha +\beta )$ is equal to

• A. 14
• B. 16
• C. 18
• D. 20

Answer 1

Answer: (D)

$\frac{x^2}{16}-\frac{y^2}{9}=1$
Let point $P(4\sec \theta ,3\tan \theta )$
Tangent at point $P$ is given by $T=0$
$\frac{x\sec \theta }{4}-\frac{y\tan \theta }{3}=1$
$A=\left(\frac{4}{\sec \theta },0\right),B=\left(0,\frac{-3}{\tan \theta }\right)$
Let mid point of $AB$ is $M(h,k)$
$\therefore h=\frac{4}{2\sec \theta },k=\frac{-3}{2\tan \theta }$
As we know, ${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow \frac{16}{4h^2}-\frac{9}{4k^2}=1\Rightarrow \frac{16}{x^2}-\frac{9}{y^2}=$
Compare with $\frac{\alpha }{x^2}-\frac{9}{y^2}=\beta ;\alpha =16,\beta =4\Rightarrow (\alpha +\beta )=20$.