Question

If the curves $\frac{x^2}{a^2}+\frac{y^2}{16}=1$ and $y^3=8x$ intersect at right angles, then $a^2$ is equal to :

• A. $\frac{16}{3}$
• B. $12$
• C. $8$
• D. $4$
• E. $2$

Answer 1

The correct option is A

$\frac{16}{3}$

Explanation for the correct option:

Step 1: Find slopes of the curves

Given, curves $\frac{x^2}{a^2}+\frac{y^2}{16}=1$ and $y^3=8x$ intersect at right angle.

For curve, $y^3=8x$.

Differentiate $y^3=8x$ with respect to $x$ to get slope.
$$\begin{gathered} \Rightarrow 3y^2\frac{dy}{dx}=8 \\ \Rightarrow m_1=\frac{8}{3y^2} \end{gathered}$$

For curve, $\frac{x^2}{a^2}+\frac{y^2}{16}=1$ slope is:
$$\begin{gathered} \Rightarrow \frac{2x}{a^2}+\frac{2y}{16}\frac{dy}{dx}=0 \\ \Rightarrow m_2=\frac{-x}{a^2}\times \frac{16}{y} \end{gathered}$$
Step 2: Find value of $a^2$

As the curves are at right angle so product of their slopes is $-1$.
$$\begin{gathered} \Rightarrow {\mathrm{m}}_1·{\mathrm{m}}_2=-1 \\ \Rightarrow \frac{8}{3{\mathrm{y}}^2}\left(\frac{-16\mathrm{x}}{{\mathrm{a}}^2\mathrm{y}}\right)=-1 \\ \Rightarrow \frac{16\times 8\times \mathrm{x}}{3{\mathrm{a}}^2{\mathrm{y}}^3}=1 \\ \Rightarrow 128\mathrm{x}=3{\mathrm{a}}^2{\mathrm{y}}^3 \\ \Rightarrow {\mathrm{a}}^2=\frac{128\mathrm{x}}{3{\mathrm{y}}^3} \\ \Rightarrow {\mathrm{a}}^2=\frac{128\mathrm{x}}{3\times 8\mathrm{x}}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{y}}^{\mathbf{3}}\mathbf{=}\mathbf{8}\mathbf{x}] \\ \Rightarrow {\mathrm{a}}^2=\frac{16}{3} \end{gathered}$$
Hence, option A is correct.