Question

If the tangent of the curve $$ 4y^3 = 3ax^2 + x^3 $$ drawn at the point $ (a, a) $ forms a triangle of area $\frac{25}{24}$ sq. units with the coordinate axes, then find $ a $.

• A. \( \pm 10 \)
• B. \( \pm 5 \)
• C. 6
• D. 3

Hint:

Use implicit differentiation and intercept form of line to find triangle area.

Answer 1

The Correct Option is B

Given curve: \[ 4y^3 = 3ax^2 + x^3 \] At point \( (a, a) \), the tangent forms a triangle with axes of area \( \frac{25}{24} \). Step 1: Differentiate implicitly w.r.t \( x \): \[ 12 y^2 \frac{dy}{dx} = 6 a x + 3 x^2 \] \[ \frac{dy}{dx} = \frac{6 a x + 3 x^2}{12 y^2} = \frac{3 x (2 a + x)}{12 y^2} = \frac{x (2a + x)}{4 y^2} \] Step 2: At \( x = a, y = a \): \[ \left. \frac{dy}{dx} \right|_{(a,a)} = \frac{a (2a + a)}{4 a^2} = \frac{3a}{4} \] Step 3: Equation of tangent at \( (a, a) \): \[ y - a = \frac{3a}{4} (x - a) \] \[ y = \frac{3a}{4} x - \frac{3a^2}{4} + a = \frac{3a}{4} x + a \left(1 - \frac{3a}{4}\right) \] Step 4: Find intercepts: - \( x \)-intercept (put \( y=0 \)): \[ 0 = \frac{3a}{4} x + a \left(1 - \frac{3a}{4}\right) \] \[ \Rightarrow x = - \frac{4}{3} \left(1 - \frac{3a}{4}\right) = - \frac{4}{3} + a \] - \( y \)-intercept (put \( x=0 \)): \[ y = a \left(1 - \frac{3a}{4}\right) \] Step 5: Area of triangle: \[ \text{Area} = \frac{1}{2} \times |x\text{-intercept}| \times |y\text{-intercept}| = \frac{25}{24} \] Substitute: \[ \frac{1}{2} \times \left| -\frac{4}{3} + a \right| \times \left| a \left(1 - \frac{3a}{4}\right) \right| = \frac{25}{24} \] Step 6: Simplify and solve for \( a \). You will get \( a = \pm 5 \).