Question

Two tangents are drawn from a point on hyperbola $x^2-y^2=5$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If they make angle $\alpha$ and $\beta$ with x-axis then:

• A. $\alpha -\beta =\pm \frac{\pi }{2}$
• B. $\alpha -\beta =\frac{\pi }{2}$
• C. $\alpha -\beta =\pi$
• D. $\alpha -\beta =0$

Answer 1

Answer: (B)

$\alpha -\beta =\frac{\pi }{2}$

Given equation of ellipse is
$\frac{x^2}{9}+\frac{y^2}{4}=1$
Here, $a^2=9,b^2=4$
Let $m$ be the slope of tangent to ellipse.
Equation of tangent to ellipse is
$y=mx+\sqrt{9m^2+4}$
Given equation of hyperbola is
$x^2-y^2=5$
Any point on the hyperbola is of the form $(\sqrt{5}\sec \theta ,\sqrt{5}\tan \theta )$
Since, the tangent passes through $(\sqrt{5}\sec \theta ,\sqrt{5}\tan \theta )$
$\sqrt{5}\tan \theta -m\sqrt{5}\sec \theta =\sqrt{9m^2+4}$
Squaring both sides, we get
$9m^2+4=5{\tan }^2\theta +5m^2{\sec }^2\theta -10m\sec \theta \tan \theta$
$\Rightarrow (9-5{\sec }^2\theta )m^2+10\tan \theta \sec \theta m+(4-5{\tan }^2\theta )=0$
$\Rightarrow (4-5{\tan }^2\theta )+10\tan \theta \sec \theta m+(4-5{\tan }^2\theta )=0$
Here, product of roots $m_1{m}_2=1$
$\Rightarrow \tan \alpha \tan \beta =1$
Now, $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
$\tan (\alpha +\beta )=\frac{m_1+m_2}{1-1}$
$\Rightarrow (\alpha +\beta )=\frac{\pi }{2}$