Mathematics - Application of Derivative Question with Solution | TestHub

$\because 2y^3=a{x}^2+x^3\Rightarrow 6y^2\frac{dy}{dx}=2ax+3x^2$ $\therefore {\left(\frac{dy}{dx}\right)}_{(a,a)}=\frac{2a^2+3a^2}{6a^2}=\frac{5}{6}$ ⇒ Equation of tangent is $y-a=\frac{5}{6}(x-a)$ or $-5x+6y=a$ or $\frac{x}{(-a/5)}+\frac{y}{(a/6)}=1$ $\begin{array}{rlr}& \therefore \alpha =-\frac{a}{5},\beta =\frac{a}{6}& {\left(-\frac{a}{5}\right)}^2+{\left(\frac{a}{6}\right)}^2=61\text{ (given) }\Rightarrow \frac{61a^2}{900}=61\Rightarrow a=\pm 30\end{array}$