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If the curves $\frac{x^2}{\alpha}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect at right angles, then a value of $\alpha$ is :
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2579 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{3}$
$\frac{4}{3}$
$$
\begin{aligned}
& \text { } \frac{x^2}{\alpha}+\frac{y^2}{4}=1 \Rightarrow \frac{2 x}{\alpha}+\frac{2 y}{4} \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-4 x}{\alpha y} \\
& y^3=16 x \Rightarrow 3 y^2 \cdot \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2} \ldots
\end{aligned}
$$
Since curves intersects at right angles
$$
\therefore \frac{-4 x}{\alpha y} \times \frac{16}{3 y^2}=-1 \Rightarrow 3 \alpha y^3=64 x
$$
\begin{aligned}
& \text { } \frac{x^2}{\alpha}+\frac{y^2}{4}=1 \Rightarrow \frac{2 x}{\alpha}+\frac{2 y}{4} \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-4 x}{\alpha y} \\
& y^3=16 x \Rightarrow 3 y^2 \cdot \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2} \ldots
\end{aligned}
$$
Since curves intersects at right angles
$$
\therefore \frac{-4 x}{\alpha y} \times \frac{16}{3 y^2}=-1 \Rightarrow 3 \alpha y^3=64 x
$$
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