If the curves frac{x^2}{a^2} + frac{y^2}{4} = | vedclass

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(C) Given curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$. Let the curves intersect at point $(x_1, y_1)$. For the first curve $\frac

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$\frac{4}{3}$

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(C) Given curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$. Let the curves intersect at point $(x_1, y_1)$. For the first curve $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$,differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{a^2y}$. So,$m_1 = -\frac{4x_1}{a^2y_1}$. For the second curve $y^3 = 16x$,differentiating with respect to $x$: $3y^2 \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{16}{3y^2}$. So,$m_2 = \frac{16}{3y_1^2}$. Since the curves intersect at right angles,$m_1 	imes m_2 = -1$. $(-\frac{4x_1}{a^2y_1}) 	imes (\frac{16}{3y_1^2}) = -1$. $\frac{64x_1}{3a^2y_1^3} = 1$. Since $(x_1, y_1)$ lies on $y^3 = 16x$,we have $y_1^3 = 16x_1$. Substituting this into the equation: $\frac{64x_1}{3a^2(16x_1)} = 1 \implies \frac{64x_1}{48a^2x_1} = 1 \implies \frac{4}{3a^2} = 1 \implies a^2 = \frac{4}{3}$.

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 =$

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