Question

If the curves $2x^2+3y^2=6$ and $a{x}^2+4y^2=4$ intersect orthogonally then $a=$

• A. $2$
• B. $1$
• C. $3$
• D. $-3$

Answer 1

The correct option is A

$2$

Explanation for the correct option.

Step 1: Find the slope of the tangents

Given curves $2x^2+3y^2=6$ and $a{x}^2+4y^2=4$ intersect orthogonally it means that their tangents are perpendicular.

The slope of the tangent to the curve, $2x^2+3y^2=6$.

Differentiate with respect to $x$ both sides we get:

$$\begin{gathered} 4x+6y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{3y}....\left(1\right) \end{gathered}$$

The slope of the tangent to the curve, $a{x}^2+4y^2=4$.

Differentiate with respect to $x$ both sides we get:

$$\begin{gathered} 2ax+8y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-ax}{4y}...\left(2\right) \end{gathered}$$

Step 2: Find the value of $a$

We know that when two lines are perpendicular then the product of their slope is $-1$.

$$\begin{gathered} \Rightarrow \left(\frac{-2x}{3y}\right)\left(\frac{-ax}{4y}\right)=-1 \\ \Rightarrow \frac{a{x}^2}{6y^2}=-1 \\ \Rightarrow \frac{x^2}{y^2}=-\frac{6}{a} \end{gathered}$$

Substitute $\frac{x^2}{y^2}=-\frac{6}{a}$in any of the given equations to get:

$a=2$

Hence, option A is correct.