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 $4 h^2=$

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 $4 h^2=$
  1. $(a+2 b)(a+3 b)$
  2. $a^2+4 a b+b^2$
  3. $a^2+6 a b+b^2$
  4. $(\mathrm{a}-2 \mathrm{~b})(2 \mathrm{a}+\mathrm{b})$

Solution

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 $\begin{aligned} & (\mathrm{a}+\mathrm{b})^2=4\left(\mathrm{~h}^2-\mathrm{ab}\right) \\ & \therefore 4 \mathrm{~h}^2=\mathrm{a}^2+6 \mathrm{ab}+\mathrm{b}^2 \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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