Question

If the normal drawn at a point $ P $ on the curve $ 3y = 6x - 5x^3 $ passes through the origin $(0, 0)$, then the positive integral value of the abscissa of point $ P $ is:

• A. 1
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{3} \)
• D. \( -\frac{2}{3} \)

Hint:

For normals passing through a point, use point-slope form with the negative reciprocal of the derivative.

Answer 1

The Correct Option is A

Given: \( 3y = 6x - 5x^3 \Rightarrow y = 2x - \frac{5}{3}x^3 \) Differentiate: \[ \frac{dy}{dx} = 2 - 5x^2 \Rightarrow \text{Slope of tangent at } x = a:\ m_T = 2 - 5a^2 \Rightarrow \text{Slope of normal } m_N = -\frac{1}{m_T} = \frac{-1}{2 - 5a^2} \] Let point \( P = (a, y(a)) \). Since the normal passes through the origin: Using point-slope form: \[ y - y_1 = m(x - x_1) \Rightarrow y - y(a) = \frac{-1}{2 - 5a^2}(x - a) \] Put \( x = 0, y = 0 \) as it passes through origin: \[ 0 - (2a - \frac{5}{3}a^3) = \frac{-1}{2 - 5a^2}(0 - a) \Rightarrow -2a + \frac{5}{3}a^3 = \frac{a}{2 - 5a^2} \] Multiply both sides by \( 2 - 5a^2 \): \[ (-2a + \frac{5}{3}a^3)(2 - 5a^2) = a \] Now try \( a = 1 \): LHS: \( (-2 + \frac{5}{3})(2 - 5) = (-\frac{1}{3})(-3) = 1 = \text{RHS} \) So, the positive integral abscissa is \( \boxed{1} \)