The angle between the lines represented by ax | vedclass

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

Question

(B) The given equation is a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$,which represents a pair of straight lines passing through the

Vedclass · Accepted Answer

$	an^{-1} \left( \frac{2\sqrt{h^2 - ab}}{a + b} ight)$

Vedclass · Answer

(B) The given equation is a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$,which represents a pair of straight lines passing through the origin.Let the slopes of the two lines be $m_1$ and $m_2$.Then,$m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.We know that $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = \left( -\frac{2h}{b} ight)^2 - 4\left( \frac{a}{b} ight) = \frac{4h^2 - 4ab}{b^2} = \frac{4(h^2 - ab)}{b^2}$.Therefore,$|m_1 - m_2| = \frac{2\sqrt{h^2 - ab}}{|b|}$.Substituting these values into the formula for $	an 	heta$:$	an 	heta = \left| \frac{\frac{2\sqrt{h^2 - ab}}{b}}{1 + \frac{a}{b}} ight| = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Thus,$	heta = 	an^{-1} \left( \frac{2\sqrt{h^2 - ab}}{a + b} ight)$.

The angle between the lines represented by $ax^2 + 2hxy + by^2 = 0$ is:

Similar Questions

If $4x^2+12xy+9y^2+2gx+2fy-1=0$ represents a pair of parallel lines,then:

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

Which of the following equations represents a pair of perpendicular straight lines?

The angle between the pair of straight lines $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

The angle between the pair of lines $2x^2 + 5xy + 2y^2 + 3x + 3y + 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module