Question

If the curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16x$ intersect at right angles, then $a^2$ is equal to

• A. $5/3$
• B. $4/3$
• C. $6/11$
• D. $3/2$

Answer 1

The correct option is B $4/3$

$y^3=16x$

$3y^2\frac{dy}{dx}=16$

$\frac{dy}{dx}=m_1=\frac{16}{3y^2}$

$\frac{x^2}{a^2}+\frac{y^2}{4}=1$

$\frac{2x}{a^2}+\frac{2y}{4}\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-x}{a^2}\times \frac{4}{y}=m_2$

$m_1.m_2=-1$

$\frac{16}{3y^2}\left(\frac{-4x}{a^{2}y}\right)=-1$

$\frac{16\times 4\times x}{3a^2{y}^3}=1\Rightarrow 64x=3a^2{y}^3$

$a^2=\frac{64x}{3y^3}$ Now $y^3=16x$

$a^2=\frac{64x}{3\times 16x}\Rightarrow a^2=\frac{4}{3}$