# then which of the following relations is true?

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 true?

1. (1) $a+b=c+d$

2. (2) $a-b=c-d$

3. (3) $a b=\frac{c+d}{a+b}$

4. (4) $a-c=b+d$

Correct Option: 2,

Solution:

$\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$

diff : $\frac{2 x}{a}+\frac{2 y}{b} \frac{d y}{d x}=0 \Rightarrow \frac{y}{b} \frac{d y}{d x}=\frac{-x}{a}$

$\frac{d y}{d x}=\frac{-b x}{a y} \ldots(2)$

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

Diff : $\frac{d y}{d x}=\frac{-d x}{c y}$....(4)

$\mathrm{m}_{1} \mathrm{~m}_{2}=-1 \Rightarrow \frac{-\mathrm{bx}}{\mathrm{ay}} \times \frac{-\mathrm{dx}}{\mathrm{cy}}=-1$

$\Rightarrow \mathrm{bdx}^{2}=-\mathrm{acy}^{2} \quad \ldots \ldots(5)$

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

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

$\Rightarrow(c-a)-(d-b)=0$

$\Rightarrow c-a=d-b$