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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 $90°$, then which of the following relations is true?

A

$a+b=c+d$

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B

$a-b=c-d$

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C

$ab=\frac{c+d}{a+b}$

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D

$a-c=b+d$

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Solution

## The correct option is B $a-b=c-d$Explanation for the correct optionStep1: Solve for slopes of the curvesThe given equation of the curves are$\frac{{x}^{2}}{a}+\frac{{y}^{2}}{b}=1_\left(\mathrm{i}\right)$$\frac{{x}^{2}}{c}+\frac{{y}^{2}}{d}=1_\left(\mathrm{ii}\right)$Differentiate both sides of the equation $\left(\mathrm{i}\right)$ with respect to $x$.$\frac{d}{dx}\left(\frac{{x}^{2}}{a}+\frac{{y}^{2}}{b}\right)=\frac{d}{dx}\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{a}+\frac{2y}{b}·\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{bx}{ay}$Thus, slope of the curve, ${m}_{1}=\frac{dy}{dx}=-\frac{bx}{ay}$.Differentiate both sides of the equation $\left(\mathrm{ii}\right)$ with respect to $x$.$\frac{d}{dx}\left(\frac{{x}^{2}}{c}+\frac{{y}^{2}}{d}\right)=\frac{d}{dx}\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{c}+\frac{2y}{d}·\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{dx}{cy}$Thus, slope of the curve, ${m}_{2}=\frac{dy}{dx}=-\frac{dx}{cy}$Step 2: Solve for the required relation.As the curves intersect with each other at an angle of $90°$.So, ${m}_{1}{m}_{2}=-1$$⇒\left(-\frac{bx}{ay}\right)\left(-\frac{dx}{cy}\right)=-1\phantom{\rule{0ex}{0ex}}⇒bd{x}^{2}=-ac{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-\frac{ac}{bd}{y}^{2}$Equation $\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$.$\left(\frac{{x}^{2}}{a}+\frac{{y}^{2}}{b}\right)-\left(\frac{{x}^{2}}{c}+\frac{{y}^{2}}{d}\right)=1-1\phantom{\rule{0ex}{0ex}}⇒\left(\frac{{x}^{2}}{a}-\frac{{x}^{2}}{c}\right)+\left(\frac{{y}^{2}}{b}-\frac{{y}^{2}}{d}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\frac{1}{a}-\frac{1}{c}\right){x}^{2}+\left(\frac{1}{b}-\frac{1}{d}\right){y}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\frac{c-a}{ac}\left(-\frac{ac}{bd}{y}^{2}\right)=-\frac{d-b}{bd}{y}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{a-c}{bd}{y}^{2}=\frac{b-d}{bd}{y}^{2}\phantom{\rule{0ex}{0ex}}⇒a-c=b-d\phantom{\rule{0ex}{0ex}}⇒a-b=c-d$Therefore, the correct relation is $a-b=c-d$.Hence, option(B) is the correct option.

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