Question

If the curves $\frac{x^2}{a^2}+\frac{y^2}{12}=1$ and $y^3=8x$ intersect at right angle, then the value of $a^2$ is equal to

• A. 16
• B. 12
• C. 8
• D. 4
• E. 2

Answer 1

Answer: (D)

Given curves are $\frac{x^2}{a^2}+\frac{y^2}{12}=1$
$\Rightarrow$ $\frac{2x}{a^2}+\frac{2y}{12}.\frac{dy}{dx}=0$
$\Rightarrow$ $\frac{dy}{dx}=-\frac{12x}{a^{2}y}=m_1$
(say) and $y^3=8x\Rightarrow 3y^2\frac{dy}{dx}=8$
$\Rightarrow$ $\frac{dy}{dx}=\frac{8}{3y^2}=m_2$ (say) For $\theta =\frac{\pi }{2},1+m_1{m}_2=0$
$\Rightarrow$ $1+\left(\frac{-12x}{a^{2}y}\right)\left(\frac{8}{3y^2}\right)=0$
$\Rightarrow$ $3a^2(8x)-96x=0$
$\Rightarrow$ $a^2=4$