If the tangent to the curve $2y^2=ax^2+a^3$ at the point (a, a) cuts off intercepts α and β on the coordinate axes such that $α^2+β^2=61$, then a =

± 30

$\frac{dy}{dx}(a,a)=\frac{2ax+3x^2}{6y^2}=\frac{5a^2}{6a^2}=\frac{5}{6}$

Equation of tangent is $y-a=\frac{5}{6}(x-a)⇒5x-6y+a=0$

$α=-a/5,\,β=a/6$

$α^2+β^2=61⇒\frac{a^2}{25}+\frac{a^2}{36}=61⇒61a^2=61(25)(36)$

$⇒a± 30$