Question

If the tangent to the curve $2y^3=a{x}^2+x^3$ at a point $(a,a)$ cuts off intercepts p and q on the coordinates axes where $p^2+q^2=61$ then $a$ equals to

• A. $30$
• B. $-30$
• C. $0$
• D. $\pm 30$

Answer 1

The correct option is D $\pm 30$

Given Curve is, $2y^3=a{x}^2+x^3$

Differentiating w.r.t $x$, we get

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

Thus slope of tangent at $(a,a)$ is $={\left(\frac{dy}{dx}\right)}_{(a,a)}=\frac{2ax+3x^2}{6y^2}=\frac{2a^2+3a^2}{6a^2}=\frac{5}{6}$

Thus equation of tangent is,

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

$\Rightarrow 6y-6a=5x-5a$

$\Rightarrow 5x-6y+a=0$

$\Rightarrow \frac{5}{-\frac{a}{5}}+\frac{y}{\frac{a}{6}}=1$

$\therefore p=\frac{-a}{5},q=\frac{a}{6}$

Given, $p^2+q^2=\frac{a^2}{25}+\frac{a^2}{36}=61$

$\Rightarrow a^2\left(\frac{61}{25\times 36}\right)=61$

$\Rightarrow a=\pm 5\times 6=\pm 30$

Hence, option 'D' is correct.