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 equal to

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

Answer 1

Answer: (C)

$\pm 30$

$2y^3=a{x}^2+x^3$

Diff. w.r.t $x$

$6y^2\frac{dy}{dx}=2ax+3x^2$

$\frac{dy}{dx}{|}_{(a,a)}=\frac{2a^2+3a^2}{6a^2}=\frac{5}{6}$

Hence tangent at $(a,a)$ is,

$y-a=\frac{5}{6}(x-a)\Rightarrow 6y-6a=5x-5a\Rightarrow 5x-6y+a=0$

$\therefore x-intercept=\frac{-a}{5}=\alpha$

and $y-intercept=\frac{a}{6}=\beta$

$\therefore {\alpha }^2+{\beta }^2=61\Rightarrow \frac{a^2}{25}+\frac{a^2}{36}=61$

$\frac{61a^2}{900}=61\Rightarrow a^2=900$

$\therefore a=\pm 30$