If $4 x-3 y-5=0$ is a normal to the ellipse $3 x^2+8 y^2=k$, then the equation of the tangent drawn to this ellipse at the point $(-2, m)(m\gt0)$ is

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Aicharya · Accepted Answer

Given the equation of ellipse $3 x^2+8 y^2=k$ $\Rightarrow \frac{x^2}{\frac{k}{3}}+\frac{y^2}{\frac{k}{8}}=1$ Now, equation of normal to the ellipse at $\left(x_1, y_1ight)$ is $\frac{a^2}{x_1} x-\frac{b^2}{y_1} y=a^2-b^2$ $\begin{aligned} & \Rightarrow \frac{k}{38 x_1} x-\frac{k}{8 y_1} y=\frac{k}{3}-\frac{k}{8} \ & \Rightarrow \frac{x}{3 x_1}-\frac{y}{8 y_1}=\frac{5}{24} \Rightarrow \frac{8}{x_1} x-\frac{3}{y_1} y=5\end{aligned}$ Compare it with $4 x-3 y-5=0$ We get $\left(x_1, y_1ight)=(2,1)$ Now, $k=3 	imes 4+8=20$ and $3 	imes 4+8 m^2=20 \Rightarrow m=1, m\gt0$ Since, $3 x^2+8 y^2=20$ $\Rightarrow 6 x+16 y y^{\prime}=0$ (Differentiate both side) $\begin{aligned} & \Rightarrow y^{\prime}=\frac{-6 x}{16 y} \ & 	ext { at }(-2,1), y^{\prime}=\frac{12}{16}=\frac{3}{4}\end{aligned}$ Now, equation of tangent at $(-2,1)$ is $\begin{aligned}& y-1=\frac{3}{4}(x+2) \Rightarrow 4 y-4=3 x+6 \ & \Rightarrow 3 x-4 y+10=0\end{aligned}$

If $4 x-3 y-5=0$ is a normal to the ellipse $3 x^2+8 y^2=k$, then the equation of the tangent drawn to this…

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