Question

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

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ...(i)
On differentiating w.r.t. $x$, we get
$ \frac{2x}{a^{2}} + \frac{2y}{b^{2}}. \frac{dy}{dx} = 0 $
$\Rightarrow \frac{dy}{dx} = - \frac{xb^{2}}{a^{2}y}$ 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)_{c_1} \times\left(\frac{dy}{dx}\right)_{c_2} = - 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} $