If the line y = mx bisects the angle between | vedclass

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

Question

(A) The equation of the bisectors of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.This can

Vedclass · Accepted Answer

$hx^2 + (a - b)x - h = 0$

Vedclass · Answer

(A) The equation of the bisectors of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.This can be rewritten as $h(x^2 - y^2) - (a - b)xy = 0 \dots (1)$.Since the line $y = mx$ is a bisector,it must satisfy equation $(1)$.Substituting $y = mx$ into equation $(1)$,we get:$h(x^2 - (mx)^2) - (a - b)x(mx) = 0$$h(x^2 - m^2x^2) - (a - b)mx^2 = 0$Dividing by $x^2$ (assuming $x 
eq 0$):$h(1 - m^2) - (a - b)m = 0$$h - hm^2 - (a - b)m = 0$$hm^2 + (a - b)m - h = 0$.Thus,$m$ is a root of the quadratic equation $hx^2 + (a - b)x - h = 0$.

If the line $y = mx$ bisects the angle between the lines $ax^2 + 2hxy + by^2 = 0$,then $m$ is a root of the quadratic equation:

Similar Questions

The equation of a pair of lines $y=px$ and $y=qx$ can be written as $(y-px)(y-qx)=0$. Then the equation of the pair of angle bisectors of the lines $x^2-4xy-5y^2=0$ is

Find the equation of the pair of straight lines that bisect the angles between the lines represented by $ax^2 + 2hxy + by^2 = 0$.

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent,then the value of $\lambda$ is

The ratio in which the line $x+y-1=0$ divides the line segment joining the origin $(0,0)$ and the point of intersection of the lines represented by $2x^2-13xy-7y^2+x+23y-6=0$ is

If $(-1, -1)$ is the point of intersection of the pair of lines $2x^2 + 5xy - 3y^2 + 2gx + 2fy + c = 0$,then $g + f =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module