If the pair of lines given by (x cos alpha + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ Expanding the left side: $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha +

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given equation: $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ Expanding the left side: $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha$ Subtracting $y^2 \sin^2 \alpha$ from both sides: $x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha$ Rearranging: $x^2(\cos^2 \alpha - \sin^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$ This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \cos^2 \alpha - \sin^2 \alpha$,$h = \sin \alpha \cos \alpha$,and $b = 0$. For the lines to be perpendicular,the condition is $a + b = 0$. Substituting the values: $(\cos^2 \alpha - \sin^2 \alpha) + 0 = 0$ $\cos^2 \alpha = \sin^2 \alpha$ $	an^2 \alpha = 1$ Since $\alpha$ is typically in the first quadrant for such problems,$\alpha = \frac{\pi}{4}$.

If the pair of lines given by $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ are perpendicular to each other,then $\alpha$ is

Similar Questions

If the pair of lines given by $(x^2+y^2) \cos^2 \theta = (x \cos \theta + y \sin \theta)^2$ are perpendicular to each other,then $\theta$ is equal to

The angle between the lines represented by the equation $y^{2} \sin^{2} \theta - xy \sin^{2} \theta + x^{2}(\cos^{2} \theta - 1) = 0$ is

The angle between the pair of lines $x^{2}+2xy-y^{2}=0$ is

If the acute angle between the lines $x^{2}-4xy+y^{2}=0$ is $\tan^{-1}(k)$,then $k=$

The angle between the lines represented by the equation ${x^2} - xy - 6{y^2} - 7x + 31y - 18 = 0$ is.....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module