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The line $x+m y+n=0$ will be a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if
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Verified Answer
The correct answer is:
$a^2 \ell^2-b^2 m^2=n^2$
If $y=M x+C$ is tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $C^2=a^2 M^2-b^2$.
$l x+m y+n=0 \Rightarrow y=-\frac{1}{m} x-\frac{n}{m}$
Comparing this equation with $\mathrm{y}=\mathrm{Mx}+\mathrm{C}$, we get
$M=-\frac{1}{m}, C=-\frac{n}{m}$
Now, $\mathrm{C}^2=\mathrm{a}^2 \mathrm{M}^2-\mathrm{b}^2$
$\begin{aligned}
& \Rightarrow \frac{n^2}{m^2}=a^2 \frac{l^2}{m^2}-b^2 \\
& \Rightarrow n^2=a^2 l^2-b^2 m^2
\end{aligned}$
$l x+m y+n=0 \Rightarrow y=-\frac{1}{m} x-\frac{n}{m}$
Comparing this equation with $\mathrm{y}=\mathrm{Mx}+\mathrm{C}$, we get
$M=-\frac{1}{m}, C=-\frac{n}{m}$
Now, $\mathrm{C}^2=\mathrm{a}^2 \mathrm{M}^2-\mathrm{b}^2$
$\begin{aligned}
& \Rightarrow \frac{n^2}{m^2}=a^2 \frac{l^2}{m^2}-b^2 \\
& \Rightarrow n^2=a^2 l^2-b^2 m^2
\end{aligned}$
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