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If $4 x+2 y+n=0$ is a normal to the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$, then $n=$
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Verified Answer
The correct answer is:
$\pm 8$
$$
\begin{aligned}
& \text 4 x+2 y+n=0 \Rightarrow y=-2 x-\frac{n}{2} \\
& \Rightarrow \frac{x^2}{36}+\frac{y^2}{16}=1
\end{aligned}
$$
If $y=m x+c$ is normal to ellipse
$$
\begin{aligned}
c^2 & =\frac{\left(a^2-b^2\right)^2 m^2}{a^2+b^2 m^2} \Rightarrow \therefore \frac{n^2}{4}=\frac{(36-16)^2(4)}{(36+16 \times 4)} \\
\Rightarrow \frac{n^2}{4} & =\frac{400 \times 4}{100} \Rightarrow n^2=64 \Rightarrow n= \pm 8 .
\end{aligned}
$$
\begin{aligned}
& \text 4 x+2 y+n=0 \Rightarrow y=-2 x-\frac{n}{2} \\
& \Rightarrow \frac{x^2}{36}+\frac{y^2}{16}=1
\end{aligned}
$$
If $y=m x+c$ is normal to ellipse
$$
\begin{aligned}
c^2 & =\frac{\left(a^2-b^2\right)^2 m^2}{a^2+b^2 m^2} \Rightarrow \therefore \frac{n^2}{4}=\frac{(36-16)^2(4)}{(36+16 \times 4)} \\
\Rightarrow \frac{n^2}{4} & =\frac{400 \times 4}{100} \Rightarrow n^2=64 \Rightarrow n= \pm 8 .
\end{aligned}
$$
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