If pairs of straight lines x^2-2 p x y-y^2=0 | vedclass

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

Question

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.For the first pa

Vedclass · Accepted Answer

$p q=-1$

Vedclass · Answer

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.For the first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.Given that this pair of bisectors is $x^2-2 q x y-y^2=0$,we compare the coefficients:Comparing $x^2+\frac{2}{p} x y-y^2=0$ with $x^2-2 q x y-y^2=0$,we get $\frac{2}{p}=-2 q$.Therefore,$p q=-1$.

If pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ be such that each pair bisects the angle between the other pair,then

Similar Questions

$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} = $

Choose the correct alternative answer

$a...be....a....beda.....ccd....bed.$

The principle of controlled chain reaction is used in

Let $\alpha \neq 1$ be a real root of the equation $x^3-ax^2+ax-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module