Question

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

• A. ${90}^0$
• B. $\alpha$
• C. $\frac{\alpha }{2}$
• D. $2\alpha$

Answer 1

The correct option is A ${90}^0$

Comparing above equation with standard $a{x}^2+2hxy+b{y}^2=0$

we get $a={\sin }^2\alpha$ and $b={\cos }^2\alpha -1$

So $a+b={\sin }^2\alpha +{\cos }^2\alpha -1=1-1=0$

Hence given line are orthogonal(perpendicular) to each other

Note: In equation of pair of line, if sum of coefficients of $x^2$ and $y^2$ is zero,

then the lines are perpendicular to each other