MAH-CET 2026 — Mathematics PYQ

1. Standard Form comparison:

The given equation of a pair of straight lines passing through the origin is:

$$y^2{\sin }^2\theta -xy{\sin }^2\theta +x^2({\cos }^2\theta -1)=0$$

Let's rearrange the terms in descending powers of $x$ to match the standard homogeneous second-degree equation form: $A{x}^2+2Hxy+B{y}^2=0$.

$$({\cos }^2\theta -1)x^2-({\sin }^2\theta )xy+({\sin }^2\theta )y^2=0$$

Using the fundamental trigonometric identity ${\sin }^2\theta +{\cos }^2\theta =1$, we can simplify the coefficient of $x^2$:

$${\cos }^2\theta -1=-{\sin }^2\theta$$

Substituting this back into the equation gives:

$$(-{\sin }^2\theta )x^2-({\sin }^2\theta )xy+({\sin }^2\theta )y^2=0$$

2. Identifying the Coefficients:

Comparing this with the general form $A{x}^2+2Hxy+B{y}^2=0$:

• Coefficient of $x^2$ ($A$) = $-{\sin }^2\theta$

• Coefficient of $xy$ ($2H$) = $-{\sin }^2\theta$

• Coefficient of $y^2$ ($B$) = ${\sin }^2\theta$

3. Finding the Angle Between the Lines:

The condition for two lines represented by a general homogeneous second-degree equation to be perpendicular ($\alpha =90^{\circ }$ or $\frac{\pi }{2}$) is:

$$\text{Coefficient of }{x}^2+\text{Coefficient of }{y}^2=0$$

$$A+B=0$$

Let's check the value of $A+B$ for our equation:

$$A+B=(-{\sin }^2\theta )+({\sin }^2\theta )=0$$

4. Conclusion:

Since the sum of the coefficients of $x^2$ and $y^2$ is exactly zero ($A+B=0$), the pair of straight lines are perpendicular to each other.

Therefore, the angle between the pair of straight lines is $\frac{\pi }{2}$.