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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, where $\alpha^{2}+\beta^{2}=61$, then the value of $a$ is equal to
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The correct answer is:
$\pm 30$
Given, equation of curve is
$2 y^{3}=a x^{2}+x^{3}$
$\Rightarrow \quad 6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}$
$\left.\Rightarrow \quad \frac{d y}{d x}\right|_{(a, a)}=\frac{2 a^{2}+3 a^{2}}{6 a^{2}}=\frac{5 a^{2}}{6 a^{2}}=\frac{5}{6}$
So, $x$-intercept $=-\frac{a}{5}=\alpha$
and $y$-intercept $=\frac{a}{6}=\beta$
$\because \quad \alpha^{2}+\beta^{2}=61$
$\Rightarrow \quad \frac{a^{2}}{25}+\frac{a^{2}}{36}=61$
$\Rightarrow \quad \frac{61 a^{2}}{36 \times 25}=61$
$\Rightarrow \quad a^{2}=36 \times 25$
$\Rightarrow \quad a=\pm 6 \times 5=\pm 30$
$2 y^{3}=a x^{2}+x^{3}$
$\Rightarrow \quad 6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}$
$\left.\Rightarrow \quad \frac{d y}{d x}\right|_{(a, a)}=\frac{2 a^{2}+3 a^{2}}{6 a^{2}}=\frac{5 a^{2}}{6 a^{2}}=\frac{5}{6}$
So, $x$-intercept $=-\frac{a}{5}=\alpha$
and $y$-intercept $=\frac{a}{6}=\beta$
$\because \quad \alpha^{2}+\beta^{2}=61$
$\Rightarrow \quad \frac{a^{2}}{25}+\frac{a^{2}}{36}=61$
$\Rightarrow \quad \frac{61 a^{2}}{36 \times 25}=61$
$\Rightarrow \quad a^{2}=36 \times 25$
$\Rightarrow \quad a=\pm 6 \times 5=\pm 30$
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