MHT CET · Maths · Pair of Lines
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 \(4 h^2=\)
- A \((a+2 b)(a+3 b)\)
- B \(a^2+4 a b+b^2\)
- C \(a^2+6 a b+b^2\)
- D \((\mathrm{a}-2 \mathrm{~b})(2 \mathrm{a}+\mathrm{b})\)
Answer & Solution
Correct Answer
(C) \(a^2+6 a b+b^2\)
Step-by-step Solution
Detailed explanation
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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