Given curve, $2 y^3=a x^2+x^3$
and $\alpha^2+\beta^2=61$
Let $A(a, a)$ be the point.
Differentiate Eq. (i) w.r.t $x$ we get
$$
\begin{aligned}
6 y^2 \frac{d y}{d x} & =2 a x+3 x^2 \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{a x}{3 y^2}+\frac{x^2}{2 y^2} \\
\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(a, a)} & =\frac{a^2}{3 a^2}+\frac{a^2}{2 a^2}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}
\end{aligned}
$$

Now, slope of line at $(a, a)=5 / 6$
Equation of line is
$$
\begin{array}{rlrl}
y-y_1 & =m\left(x-x_1\right) \\
\Rightarrow & y-a & =\frac{5}{6}(x-a) \text { or } 6 y-5 x=a \\
\Rightarrow \quad & \frac{y}{a / 6}-\frac{x}{a / 5} & =1 \text { or } \frac{x}{-a / 5}+\frac{y}{a / 6}=1
\end{array}
$$

Compare with intercept form of line $\frac{x}{\alpha}+\frac{y}{\beta}=1$
Gives $\alpha=-\frac{a}{5}, \beta=\frac{a}{6}$
Given, $\alpha^2+\beta^2=61$
$$
\begin{aligned}
\Rightarrow & =\frac{a^2}{25}+\frac{a^2}{36} \\
& =\frac{36 a^2+25 a^2}{25 \times 36}=\frac{61 a^2}{25 \times 36}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{61 a^2}{25 \times 36}=61 \Rightarrow a^2=25 \times 36=900
$$
$$
\begin{aligned}
a & =\sqrt{900} \\
|a| & =30
\end{aligned}
$$