Question

If the tangent to the curve \(2y^3 = x^3+ax^2\) at the point \((a,a)\) cuts off intercepts \(α\) and \(β\) on the coordinate axes such that \(α^2+β^2 = 61\), then the value of \(a\) is

• A. \(\pm 61\)
• B. \(\pm 36\)
• C. \(\pm 30\)
• D. \(\pm 25\)

Hint:

Find the tangent slope at \((a,a)\) by implicit differentiation, then compute both intercepts.

Answer 1

The Correct Option is C

Step 1: Check the point:
At \((a,a)\): \(2a^3 = a^3 + a^3\). The point lies on the curve.

Step 2: Slope:
Differentiate \(2y^3 = x^3 + ax^2\): \(6y^2\frac{dy}{dx} = 3x^2 + 2ax\).
At \((a,a)\): \(6a^2 m = 3a^2 + 2a^2 = 5a^2\), so \(m = \frac56\).

Step 3: Tangent and intercepts:
Tangent: \(y - a = \frac56(x - a)\).
Put \(y = 0\): \(-a = \frac56(x-a)\), so \(x = a - \frac{6a}{5} = -\frac{a}{5}\). So \(\alpha = -\frac{a}{5}\).
Put \(x = 0\): \(y = a - \frac{5a}{6} = \frac{a}{6}\). So \(\beta = \frac{a}{6}\).

Step 4: Use the condition:
\[ \alpha^2 + \beta^2 = \frac{a^2}{25} + \frac{a^2}{36} = \frac{61a^2}{900} = 61 \]
So \(a^2 = 900\), \(a = \pm30\).

Final Answer:
The value of \(a\) is \(\pm30\), option (C). \[ \boxed{a=\pm30} \]