Question

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

• A. $16$
• B. $12$
• C. $8$
• D. $4$
• E. $2$

Answer 1

The correct option is D

$4$

Explanation for the correct option.

Step 1: Find the slopes of tangent

Given, curves $\frac{x^2}{a^2}+\frac{y^2}{12}=1$ and $y^3=8x$ intersect at right angles.

When two curves are at the right angle it means that their tangents are at a right angle.

For the slope of the tangent to the curve $\frac{x^2}{a^2}+\frac{y^2}{12}=1$ is:

$\begin{array}{rcl}\frac{2x}{a^2}+\frac{2y}{12}\frac{dy}{dx}& =& 0\\ m_1& =& \frac{-x}{a^2}\times \frac{12}{y}\end{array}$

For the slope of the tangent to the curve $y^3=8x$ is:

$\begin{array}{rcl}3y^2\frac{dy}{dx}& =& 8\\ m_2& =& \frac{8}{3y^2}\end{array}$

Step 2: Find value of $a^2$

$$\begin{gathered} \Rightarrow {\mathrm{m}}_1·{\mathrm{m}}_2=-1 \\ \Rightarrow \frac{8}{3{\mathrm{y}}^2}\left(\frac{-12\mathrm{x}}{{\mathrm{a}}^2\mathrm{y}}\right)=-1 \\ \Rightarrow \frac{32\mathrm{x}}{{\mathrm{a}}^2{\mathrm{y}}^3}=1 \\ \Rightarrow 32\mathrm{x}={\mathrm{a}}^2{\mathrm{y}}^3 \\ \Rightarrow {\mathrm{a}}^2=\frac{32\mathrm{x}}{{\mathrm{y}}^3}[\because {\mathrm{y}}^3=8\mathrm{x}] \\ \Rightarrow {\mathrm{a}}^2=\frac{32\mathrm{x}}{8\mathrm{x}} \\ \Rightarrow {\mathrm{a}}^2=4 \end{gathered}$$

Hence, option D is correct.