The angle between lines represented by $\left(\sin ^2 \alpha\right) y^2-2 x y\left(\cos ^2…
The angle between lines represented by $\left(\sin ^2 \alpha\right) y^2-2 x y\left(\cos ^2 \alpha\right)+\left(\cos ^2 \alpha-1\right) x^2=0$ is
- $2 \alpha$
- $\alpha$
- $90^{\circ}$
- $45^{\circ}$
Solution
Pair of straight lines
$
\left(\sin ^2 \alpha\right) y^2-2 x y \cos ^2 \alpha+\left(\cos ^2 \alpha-1\right) x^2=0
$
Standard form
$
\begin{aligned}
& a x^2+2 h x y+b y^2=0 \\
& \therefore \quad \begin{array}{l}
a=\cos ^2 \alpha-1 \\
h=-\cos ^2 \alpha \\
b=\sin ^2 \alpha
\end{array} \\
& \because \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|
\end{aligned}
$
$\begin{gathered}\tan \theta=\left|\frac{2 \sqrt{\cos ^4 \alpha-\left(\cos ^2 \alpha-1\right) \sin ^2 \alpha}}{\cos ^2 \alpha-1+\sin ^2 \alpha}\right| \\ \because \sin ^2 \alpha+\cos ^2 \alpha=1 \\ \therefore \quad \tan \theta=\left|\frac{2 \sqrt{\cos ^4 \alpha+\sin ^4 \alpha}}{0}\right| \\ \tan \theta=\infty \Rightarrow \theta=90^{\circ}\end{gathered}$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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