Question

At what point of the curve \( y^2 = 2x^3 \) is the tangent line perpendicular to the straight line \[ 4x - 3y + 2 = 0? \]

• A. \( \left(\frac{1}{8}, \pm \frac{1}{16} \right) \)
• B. \( \left(\frac{1}{4}, -\frac{1}{8} \right) \)
• C. \( \left(-\frac{1}{16}, \frac{1}{8} \right) \)
• D. None of these

Answer 1

The Correct Option is A

Slope of given line: \(y = \frac{4}{3}x + \frac{2}{3}\), so slope = \(\frac{4}{3}\) Required slope of tangent = negative reciprocal = \(-\frac{3}{4}\) Differentiate \( y^2 = 2x^3 \): \[ 2y \frac{dy}{dx} = 6x^2 \Rightarrow \frac{dy}{dx} = \frac{3x^2}{y} \Rightarrow \frac{3x^2}{y} = -\frac{3}{4} \Rightarrow x^2 = -\frac{y}{4} \Rightarrow y = -4x^2 \] Substitute in \( y^2 = 2x^3 \): \[ ( -4x^2 )^2 = 2x^3 \Rightarrow 16x^4 = 2x^3 \Rightarrow x = \frac{1}{8} \Rightarrow y = \pm \frac{1}{16} \]