If the bisectors of the pair of lines x^2-2 m | vedclass

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

Question

(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of

Vedclass · Accepted Answer

$mn+1=0$

Vedclass · Answer

(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$. Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{x y}{-m}$. This simplifies to $\frac{x^2-y^2}{2} = \frac{x y}{-m}$,which implies $-m(x^2-y^2) = 2 x y$. Rearranging the terms,we get $m x^2 + 2 x y - m y^2 = 0$. Dividing by $m$ (assuming $m 
eq 0$),we get $x^2 + \frac{2}{m} x y - y^2 = 0$. Comparing this with the given bisector equation $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$. Therefore,$-n = \frac{1}{m}$,which gives $mn = -1$ or $mn+1=0$.

If the bisectors of the pair of lines $x^2-2 m x y-y^2=0$ are represented by $x^2-2 n x y-y^2=0$,then

Similar Questions

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

The equation of the bisectors of the angles between the lines represented by ${x^2} + 2xy \cot \theta + {y^2} = 0$ is

If the equation $kx^2 - 5xy + 6y^2 + x - 3y = 0$ represents a pair of straight lines, then the coordinates of their point of intersection are

If the pair of straight lines $xy - x - y + 1 = 0$ and the line $ax + 2y - 3 = 0$ are concurrent,then $a =$

Let the equation of the pair of lines,$y=px$ and $y=qx$,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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module