Question

If the curves \(y^2 = 16x \text{ and } 9x^2 + \alpha y^2 = 25\) intersect at right angles, then \(\alpha =\)

• A. 6
• B. 9
• C. \(\dfrac{9}{2}\)
• D. 3

Hint:

When curves intersect orthogonally, equate the negative product of their slopes to -1 and substitute known points.

Answer 1

The Correct Option is C

Two curves intersect orthogonally if product of slopes at point of intersection is \(-1\) Let’s find derivative of: \[ y^2 = 16x \Rightarrow 2y \dfrac{dy}{dx} = 16 \Rightarrow \dfrac{dy}{dx} = \dfrac{8}{y} \] Next: \(9x^2 + \alpha y^2 = 25\) \[ \Rightarrow 18x + 2\alpha y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{9x}{\alpha y} \] Now multiply slopes: \[ \dfrac{8}{y} \cdot \left(-\dfrac{9x}{\alpha y} \right) = -\dfrac{72x}{\alpha y^2} = -1 \Rightarrow \dfrac{72x}{\alpha y^2} = 1 \] Use point of intersection: try \(x = 1, y = 4\) (satisfies \(16x = y^2\)) \[ \Rightarrow \dfrac{72 \cdot 1}{\alpha \cdot 16} = 1 \Rightarrow \alpha = \dfrac{72}{16} = \dfrac{9}{2} \]