If the tangent to the curve $2 y^3=a x^2+x^3$ at the point (a, a) cuts off intercepts $\alpha$ and $\beta$ on the coordinate axes such that $\alpha^2+\beta^2=61$, then a =
Answer & explanation
Correct answer: option 1
We have,
$2 y^3=a x^2+x^3$
$\Rightarrow 6 y^2 \frac{d y}{d x}=2 a x+3 x^2 \Rightarrow \frac{d y}{d x}=\frac{2 a x+3 x^2}{6 y^2} \Rightarrow\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{6}$
The equation of the tangent at (a, a) is
$y-a=\frac{5}{6}(x-a) \Rightarrow 5 x-6 y+a=0$
This intercepts lengths $-a / 5$ and $a / 6$ with x and y-axes respectively.
∴ $\alpha=-a / 5$ and $\beta=a / 6$
Now,
$\alpha^2+\beta^2=61 \Rightarrow \frac{a^2}{25}+\frac{a^2}{36}=61 \Rightarrow a^2=25 \times 36 \Rightarrow a= \pm 30$