Question

The angle between the lines $\sin^2 \alpha \cdot y^2 - 2xy \cdot \cos^2 \alpha + (\cos^2 \alpha - 1)x^2 = 0$, is

• A. $90^\circ$
• B. $\alpha$
• C. $\frac {\alpha}{2}$
• D. $2\alpha$

Answer 1

Answer: (A)

$90^\circ$

Answer 2

Given pair of straight line is
$\left(\cos ^{2} \alpha-1\right) x^{2}-2 \cos ^{2} \alpha \cdot x y+\sin ^{2} \alpha y^{2}=0$
On comparing with the equation
$a x^{2}+2 h \,x y+b y^{2}=0,$ we get
$a=\cos ^{2} \alpha-1=-\sin ^{2} \alpha$
$n=-\cos ^{2} \alpha \quad \text { and } \quad b=\sin ^{2} \alpha$
Let $\theta$ be the angle between the lines, then
$\tan \theta=\left|\frac{2 \sqrt{n^{2}-a b}}{a+b}\right|$
$\Rightarrow \tan \theta=\left|\frac{2 \sqrt{\cos ^{4} \alpha+\sin ^{4} \alpha}}{-\sin ^{2} \alpha+\sin ^{2} \alpha}\right|=\infty=\tan 90^{\circ}$
$\Rightarrow \theta =90^{\circ}$