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 bisectors of the angles between the pair of lines 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 equation of the bisectors $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$. This implies $mn = -1$,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

Osteomalacia is caused by the deficiency of . . . . . . .

The difference between volume and pressure coefficients of an ideal gas is

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

How are the following compounds related?

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

Vedclass Test Series

Exam Paper Generator

Online Exam Module