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 square of the distance between the origin and the point of intersection of the lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is

If the pair of lines $ax^2-7xy-3y^2=0$ and $2x^2+xy-6y^2=0$ have exactly one line in common and $a$ is an integer,then the equation of the pair of bisectors of the angles between the lines $ax^2-7xy-3y^2=0$ is

The locus of the points equidistant from the lines represented by $x^2 \cos^2 \theta - xy \sin^2 \theta - y^2 \sin^2 \theta = 0$ is

If $(2,-1)$ is the point of intersection of the pair of lines $2x^2+axy+3y^2+bx+cy-3=0$,then $3a+2b+c=$

If the equation $3 x^2-k x y-3 y^2=0$ represents the bisectors of the angles between the lines $x^2-3 x y-4 y^2=0$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module