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

Answer 1

The correct option is B

$a-b=c-d$

Explanation for the correct option

Step1: Solve for slopes of the curves

The 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 $x$.

$$\begin{gathered} \frac{d}{dx}\left(\frac{x^2}{a}+\frac{y^2}{b}\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow \frac{2x}{a}+\frac{2y}{b}·\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{bx}{ay} \end{gathered}$$

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 $x$.

$$\begin{gathered} \frac{d}{dx}\left(\frac{x^2}{c}+\frac{y^2}{d}\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow \frac{2x}{c}+\frac{2y}{d}·\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{dx}{cy} \end{gathered}$$

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, $m_1{m}_2=-1$

$$\begin{gathered} \Rightarrow \left(-\frac{bx}{ay}\right)\left(-\frac{dx}{cy}\right)=-1 \\ \Rightarrow bd{x}^2=-ac{y}^2 \\ \Rightarrow x^2=-\frac{ac}{bd}{y}^2 \end{gathered}$$

Equation $\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$.

$$\begin{gathered} \left(\frac{x^2}{a}+\frac{y^2}{b}\right)-\left(\frac{x^2}{c}+\frac{y^2}{d}\right)=1-1 \\ \Rightarrow \left(\frac{x^2}{a}-\frac{x^2}{c}\right)+\left(\frac{y^2}{b}-\frac{y^2}{d}\right)=0 \\ \Rightarrow \left(\frac{1}{a}-\frac{1}{c}\right)x^2+\left(\frac{1}{b}-\frac{1}{d}\right)y^2=0 \\ \Rightarrow \frac{c-a}{ac}\left(-\frac{ac}{bd}{y}^2\right)=-\frac{d-b}{bd}{y}^2 \\ \Rightarrow \frac{a-c}{bd}{y}^2=\frac{b-d}{bd}{y}^2 \\ \Rightarrow a-c=b-d \\ \Rightarrow a-b=c-d \end{gathered}$$

Therefore, the correct relation is $a-b=c-d$.

Hence, option(B) is the correct option.