Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ 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)

$a^2-b^2=2c^2$

Given $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .....(i)
On differentiating w.r.t. x, we get
$\frac{2x}{a^2}+\frac{2y}{b^2}.\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{x{b}^2}{a^{2}y}$

And $x^2-y^2=c^2$
On differentiating w.r.t. x, we get
$2x-2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{x}{y}$
The two curves will cut at right angles, if
${\left(\frac{dy}{dx}\right)}_{c1}\times {\left(\frac{dy}{dx}\right)}_{c2}=-1$

$\Rightarrow -\frac{b^{2}x}{a^{2}y}.\frac{x}{y}=-1$

$\Rightarrow \frac{x^2}{a^2}=\frac{y^2}{b^2}$
$\Rightarrow \frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{1}{2}$
[using eq. (i)]
On substituting these values in $x^2-y^2=c^2$, we get
$\frac{a^2}{2}-\frac{b^2}{2}=c^2$
$\Rightarrow a^2-b^2=2c^2$

Hence, option C is correct.