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If the acute angle between the lines given by $a x^2+2 h x y+b y^2=0$ is $\frac{\pi}{4}$, then $4 h^2=$
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Verified Answer
The correct answer is:
$a^2+6 a b+b^2$
As per data given we write
$$
\tan \frac{\pi}{4}=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=1
$$
Squaring both sides, we get
$$
\begin{aligned}
& (\mathrm{a}+\mathrm{b})^2=4\left(\mathrm{~h}^2-\mathrm{ab}\right) \\
& \therefore 4 \mathrm{~h}^2=\mathrm{a}^2+6 \mathrm{ab}+\mathrm{b}^2
\end{aligned}
$$
$$
\tan \frac{\pi}{4}=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=1
$$
Squaring both sides, we get
$$
\begin{aligned}
& (\mathrm{a}+\mathrm{b})^2=4\left(\mathrm{~h}^2-\mathrm{ab}\right) \\
& \therefore 4 \mathrm{~h}^2=\mathrm{a}^2+6 \mathrm{ab}+\mathrm{b}^2
\end{aligned}
$$
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