Question

lf 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 such that ${\alpha }^2+{\beta }^2=61$, then $a$ is equal to

• A. $\pm 30$
• B. $\pm 5$
• C. $\pm 6$
• D. $\pm 61$

Answer 1

The correct option is A $\pm 30$

$2\times 3\times y^2{y}^{\prime }=2ax+3x^2$

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

$m={y^{\prime }|}_{(a,a)}=\frac{5}{6}$

Equation of the tangent at $(a,a)$ with slope, $m=\frac{5}{6}$
$(y-a)=\frac{5}{6}(x-a)$

Intersection points with coordinate axes are, $(0,\frac{a}{6})$ and $(-\frac{a}{5},0)$

So, $\alpha =\frac{a}{6}$ and $\beta =-\frac{a}{5}$

Given that,

${\alpha }^2+{\beta }^2=61$

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

$\Rightarrow a^2={30}^2$

$\Rightarrow a=\pm 30$