The curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect each other orthogonally, then $\mathrm{a}^2=$

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{4}=1 \ & 	herefore \frac{1}{a^2} 2 x+\frac{1}{4} 2 y \frac{d y}{d x}=0 \quad \Rightarrow \frac{d y}{d x}=\left(\frac{-x}{a^2}ight)\left(\frac{4}{y}ight) \end{aligned}$ Also $\mathrm{y}^3=16 \mathrm{x}$ $	herefore 3 y^2 \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}$ Since curves intersect orthogonally, from (1) and (2), we write $\left(\frac{-x}{a^2}ight)\left(\frac{4}{y}ight)\left(\frac{16}{3 y^2}ight)=-1$ $	herefore \frac{64 x}{3 a^2 y^3}=1$ and we have $y^3=16 x$ $	herefore \frac{64 \mathrm{x}}{3 \mathrm{a}^2(16 \mathrm{x})}=1 \Rightarrow \mathrm{a}^2=\frac{4}{3}$

The curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect each other orthogonally, then…

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