If the curve x2α+y24=1 and y3=16x intersect at right angles then a value of 3αis _______

x2α+y24=1, y3=16xdydx=−2xα2y4=−4xαy(m1)3y2dydx=16, dydx=163y2(m2)m1`.m2=−1⇒(−4xαy)(163y2)=−164x=3αy364x=3α(16x)3α=4

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If the curve $\frac{x^{2}}{α} + \frac{y^{2}}{4} = 1$ and $y^{3} = 16 x$ intersect at right angles then a value of $3 α$ is _______

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answer is 4.

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$\frac{x^{2}}{α} + \frac{y^{2}}{4} = 1, y^{3} = 16 x$

$\frac{d y}{d x} = \frac{\frac{- 2 x}{α}}{\frac{2 y}{4}} = \frac{- 4 x}{α y} (m_{1})$

$3 y^{2} \frac{d y}{d x} = 16$ , $\frac{d y}{d x} = \frac{16}{3 y^{2}} (m_{2})$

$m_{1 `} . m_{2} = - 1 \Rightarrow (- \frac{4 x}{α y}) (\frac{16}{3 y^{2}}) = - 1$

$64 x = 3 α y^{3}$

$64 x = 3 α (16 x)$

$3 α = 4$

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