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

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(B) Given curves are $C_1: \frac{x^2}{\alpha} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.Differentiating $C_1$ with respect to $x$: $\frac{2x}{\alpha}

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$\frac{4}{3}$

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(B) Given curves are $C_1: \frac{x^2}{\alpha} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.Differentiating $C_1$ with respect to $x$: $\frac{2x}{\alpha} + \frac{2y}{4} \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{\alpha y} = m_1$.Differentiating $C_2$ with respect to $x$: $3y^2 \cdot \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{3y^2} = m_2$.Since the curves intersect at right angles,$m_1 \cdot m_2 = -1$.Substituting the slopes: $\left( -\frac{4x}{\alpha y} ight) \cdot \left( \frac{16}{3y^2} ight) = -1$.This simplifies to $\frac{64x}{3\alpha y^3} = 1 \Rightarrow 3\alpha y^3 = 64x$.Substitute $y^3 = 16x$ into the equation: $3\alpha (16x) = 64x$.Assuming $x 
eq 0$,we get $48\alpha = 64 \Rightarrow \alpha = \frac{64}{48} = \frac{4}{3}$.

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

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