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

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(B) Let the point of intersection of the two curves be $(x_1, y_1)$.The equations of the curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 1

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(B) Let the point of intersection of the two curves be $(x_1, y_1)$.The equations of the curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$.Differentiating both equations with respect to $x$:For $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$,we get $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{a^2y}$.For $y^3 = 16x$,we get $3y^2 \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{3y^2}$.Let $m_1$ and $m_2$ be the slopes of the tangents at $(x_1, y_1)$.$m_1 = -\frac{4x_1}{a^2y_1}$ and $m_2 = \frac{16}{3y_1^2}$.Since the curves intersect at right angles,$m_1 m_2 = -1$.$(-\frac{4x_1}{a^2y_1}) 	imes (\frac{16}{3y_1^2}) = -1$.$\frac{64x_1}{3a^2y_1^3} = 1$.Since $(x_1, y_1)$ lies on $y^3 = 16x$,we have $y_1^3 = 16x_1$.Substituting this into the equation: $\frac{64x_1}{3a^2(16x_1)} = 1$.$\frac{64x_1}{48a^2x_1} = 1 \Rightarrow \frac{4}{3a^2} = 1$.Therefore,$a^2 = 4/3$.

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

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