If the two lines given by ax^2+2hxy+by^2=0 ma | vedclass

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

Question

(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.From the propert

Vedclass · Accepted Answer

$\frac{2h}{a-b}$

Vedclass · Answer

(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.From the properties of the homogeneous equation of the second degree,we have:$m_1+m_2 = 	an \alpha + 	an \beta = -\frac{2h}{b}$$m_1m_2 = 	an \alpha 	an \beta = \frac{a}{b}$Using the formula for $	an(\alpha+\beta)$:$	an(\alpha+\beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$Substituting the values:$	an(\alpha+\beta) = \frac{-\frac{2h}{b}}{1 - \frac{a}{b}} = \frac{-\frac{2h}{b}}{\frac{b-a}{b}} = \frac{-2h}{b-a} = \frac{2h}{a-b}$

If the two lines given by $ax^2+2hxy+by^2=0$ make inclinations $\alpha$ and $\beta$ with the $x$-axis,then $\tan(\alpha+\beta)=$

Similar Questions

If an equation $hxy + gx + fy + c = 0$ represents a pair of lines,then

The equation of the lines passing through the origin and having slopes $3$ and $-\frac{1}{3}$ is

If one of the lines represented by $ax^2+2hxy+by^2=0$ passes through $(2,3)$ and the other passes through $(4,5)$,then $a+2h+b$ equals

If the equation $ax^2 + 2hxy + by^2 = 0$ represents two lines $y = m_1x$ and $y = m_2x$,then

The area of the triangle formed by the lines represented by $4x^2 - 9xy - 9y^2 = 0$ and the line $x = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module