Question

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

• A. $25$
• B. $36$
• C. $\pm\,30$
• D. $\pm\, 40$

Answer 1

The Correct Option is C

$2 y^{3}=a x^{2}+x^{3} $
$6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2} $
$\left.\frac{d y}{d x}\right|_{(a, a)}=\frac{5 a^{2}}{6 a^{2}}=\frac{5}{6}$
Tangent at $(a, a)$ is $5 x-6 y=-a$
$\alpha=\frac{-a}{5}, $
$\beta=\frac{a}{6}$
$\alpha^{2}+\beta^{2}=61 $
$\Rightarrow \frac{a^{2}}{25}+\frac{a^{2}}{36}=61 $
$a ^{2}=25.36 $
$ a =\pm 30$