The line x - 2y = 0 will be a bisector of the | vedclass

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

Question

(C) The equation of the pair of lines is ${x^2} - 2hxy - 2{y^2} = 0$. Comparing this with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 1$,$2h' = -2h$,and $b

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) The equation of the pair of lines is ${x^2} - 2hxy - 2{y^2} = 0$. Comparing this with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 1$,$2h' = -2h$,and $b = -2$. The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h'}$. Substituting the values,we get $\frac{x^2 - y^2}{1 - (-2)} = \frac{xy}{-h} \Rightarrow \frac{x^2 - y^2}{3} = \frac{xy}{-h}$. This simplifies to $-h(x^2 - y^2) = 3xy$,or $hx^2 + 3xy - hy^2 = 0$. Since $x - 2y = 0$ is a bisector,it must satisfy this equation. Substituting $x = 2y$,we get $h(2y)^2 + 3(2y)y - hy^2 = 0$. $4hy^2 + 6y^2 - hy^2 = 0 \Rightarrow 3hy^2 + 6y^2 = 0$. $3y^2(h + 2) = 0$. For this to hold for all $y$,$h + 2 = 0$,so $h = -2$.

The line $x - 2y = 0$ will be a bisector of the angle between the lines represented by the equation ${x^2} - 2hxy - 2{y^2} = 0$,if $h = $

Similar Questions

If the bisectors of the lines $x^2 - 2pxy - y^2 = 0$ are $x^2 - 2qxy - y^2 = 0$,then

If the equation $16x^2 - 24xy + 9y^2 - 8x + 6y - 35 = 0$ represents a pair of straight lines, then the equation of the locus of points equidistant from these two lines is...

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

If the pair of straight lines $xy-x+y-1=0$ and the line $x+ky-3=0$ are concurrent,then the value of $k$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module