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=

A
± 30
B
± 5
C
± 6
D
± 61

Solution:

We have,

$\begin{array}{l} 2 y^{3} = a x^{2} + x^{3} \\ \Rightarrow 6 y^{2} \frac{d y}{d x} = 2 a x + 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{2 a x + 3 x^{2}}{6 y^{2}} \Rightarrow {(\frac{d y}{d x})}_{(a, a)} = \frac{5}{6} \end{array}$

The equation of the tangent at $(a, a)$ is

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

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

$∴ α = - a / 5 and β = a / 6$

Now,

$α^{2} + β^{2} = 61 \Rightarrow \frac{a^{2}}{25} + \frac{a^{2}}{36} = 61 \Rightarrow a^{2} = 25 \times 36 \Rightarrow a = \pm 30$

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

Solution:

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