Question

If the tangent to the curve $2y^3=a{x}^2+x^3$ 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

Answer: (C)

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