If curves frac{x^2}{a^2} + frac{y^2}{4} = 1 a | vedclass

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(B) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.For $C_2$,differentiating with respect to $x$: $3y^2 \frac{dy}{dx

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$\frac{4}{3}$

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(B) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.For $C_2$,differentiating with respect to $x$: $3y^2 \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = m_1 = \frac{16}{3y^2}$.For $C_1$,differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = -\frac{4x}{a^2y}$.Since the curves intersect orthogonally,$m_1 	imes m_2 = -1$.$\left(\frac{16}{3y^2}ight) 	imes \left(-\frac{4x}{a^2y}ight) = -1$.$\frac{64x}{3a^2y^3} = 1 \implies 64x = 3a^2y^3$.Substitute $y^3 = 16x$ into the equation: $64x = 3a^2(16x)$.Assuming $x 
eq 0$,we get $64 = 48a^2$.$a^2 = \frac{64}{48} = \frac{4}{3}$.

If curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect orthogonally,then $a^2$ equals:

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