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: Plan:
Write the tangent in intercept form directly.

Step 2: Steps:
Slope $= \frac56$ (as $6a^2 m = 5a^2$). The tangent is $5x - 6y + c = 0$. Through $(a,a)$: $5a - 6a + c = 0$, so $c = a$. The line is $5x - 6y + a = 0$.
Intercepts: $x = -\frac{a}{5}$ (when $y=0$) and $y = \frac{a}{6}$ (when $x=0$).
$\frac{a^2}{25}+\frac{a^2}{36} = 61$ means $a^2\cdot\frac{36+25}{900} = 61$, so $a^2 = 900$ and $a = \pm 30$.

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