Question

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ and $x^2-y^2=c^2$ cut at right angles, then

• A. $a^2+b^2=2c^2$
• B. $b^2-a^2=2c^2$
• C. $a^2-b^2=2c^2$
• D. $a^2{b}^2=2c^2$

Answer 1

Answer: (C)

Let $P(a\cos \theta ,b\sin \theta )$ be a point on the ellipse.
Equation of tangent at $P$ on ellipse is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ ...(1)
equation of tangent at $P$ on $x^2-y^2=c^2$ is
$x(a\cos \theta )-y(b\sin \theta )=c^2$ ...(2)
(1) and (2) are perpendicular
$\left(-\frac{b}{a}\cot \theta \right)\cdot \left(\frac{a}{b}\cot \theta \right)=-1$
$\Rightarrow {\tan }^2\theta =1$
$P(a\cos \theta ,b\sin \theta )$ lies on the hyperbola
$x^2-y^2=c^2$
$a^2{\cos }^2\theta -b^2{\sin }^2\theta =c^2$
divide by ${\cos }^2\theta$
$a^2-b^2{\tan }^2\theta =c^2\left(1+{\tan }^2\theta \right)$
$a^2-b^2=2c^2$