If the tangent to the curve 2y3=ax2+x3 at the point (a, a) cuts off intercepts α and β on the coordinate axes such that α2+β2=61, then a=

If the tangent to the curve $2{y}^{3}=a{x}^{2}+{x}^{3}$ at the point  cuts off intercepts $\alpha$ and $\beta$ on the coordinate axes such that ${\alpha }^{2}+{\beta }^{2}=61$, then $a=$

1. A

± 30

2. B

± 5

3. C

± 6

4. D

± 61

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Solution:

We have,

The equation of the tangent at  is

$y-a=\frac{5}{6}\left(x-a\right)⇒5x-6y+a=0$

This intercepts lengths $-a/5$ and $a/6$ with $x$ and $y$-axes respectively.

Now,

${\alpha }^{2}+{\beta }^{2}=61⇒\frac{{a}^{2}}{25}+\frac{{a}^{2}}{36}=61⇒{a}^{2}=25×36⇒a=±30$

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