Question

The curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect each other orthogonally, then $a^2 =$

• A. 2
• B. 3/4
• C. 1/2
• D. 4/3

Hint:

To solve intersection problems, always express one variable from the second equation and substitute into the first or the derivative condition.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Two curves intersect orthogonally if the product of their slopes at the point of intersection is $-1$. We need to find $a^2$.

Step 2: Key Formula or Approach:
1. Differentiate both curves to find $m_1 = \frac{dy}{dx}$ and $m_2 = \frac{dy}{dx}$.
2. Set $m_1 \cdot m_2 = -1$.

Step 3: Detailed Explanation:
Curve 1: $\frac{x^2}{a^2} + \frac{y^2}{4} = 1 \implies \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \implies m_1 = -\frac{4x}{a^2 y}$.
Curve 2: $y^3 = 16x \implies 3y^2 \frac{dy}{dx} = 16 \implies m_2 = \frac{16}{3y^2}$.
Orthogonal condition: $m_1 \cdot m_2 = -1 \implies \left(-\frac{4x}{a^2 y}\right) \left(\frac{16}{3y^2}\right) = -1$.
$\frac{64x}{3a^2 y^3} = 1$. Substitute $y^3 = 16x$:
$\frac{64x}{3a^2 (16x)} = 1 \implies \frac{4}{3a^2} = 1 \implies a^2 = 4/3$.

Step 4: Final Answer:
The value is $a^2 = 4/3$, corresponding to option (D).