Question

If the curves, $\frac{x^2}{a}+\frac{y^2}{b}=1$ and $\frac{x^2}{c}+\frac{y^2}{d}=1$ intersect each other at an angle of ${90}^{\circ }$, then which of the following relations is $\text{TRUE}?$

• A. $a+b=c+d$
• B. $a-b=c-d$
• C. $ab=\frac{c+d}{a+b}$
• D. $a-c=b+d$

Answer 1

The correct option is B $a-b=c-d$

$\frac{x^2}{a}+\frac{y^2}{b}=1....\left(1\right)$
differentiate :$\frac{2x}{a}+\frac{2y}{b}\frac{dy}{dx}=0\Rightarrow \frac{y}{b}\frac{dy}{dx}=\frac{-x}{a}$

$\frac{dy}{dx}=\frac{-bx}{ay}....\left(2\right)$

$\frac{x^2}{c}+\frac{y^2}{d}=1....\left(3\right)$

Differentiate : $\frac{dy}{dx}=\frac{-dx}{cy}....\left(4\right)$

$m_1{m}_2=-1\Rightarrow \frac{-bx}{ay}\times \frac{-dx}{cy}=-1$

$\Rightarrow bd{x}^2=-ac{y}^2....\left(5\right)$

From equation$\left(1\right)-\left(3\right)\Rightarrow (\frac{1}{a}-\frac{1}{c})x^2+(\frac{1}{b}-\frac{1}{d})y^2=0$

$\Rightarrow \frac{c-a}{ac}{x}^2+\frac{d-b}{bd}\times \left(\frac{-bd}{ac}\right)x^2=0$ $($Using $5)$
$\Rightarrow (c-a)-(d-b)=0$

$\Rightarrow c-a=d-b$
$\Rightarrow c-d=a-b$