Find the angle between the lines represented | vedclass

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

Question

(A) Let the angle between the lines represented by $x^2 - 2pxy + y^2 = 0$ be $	heta$.Comparing the given equation with $ax^2 + 2hxy + by^2 = 0$,we ge

Vedclass · Accepted Answer

$sec^{-1}(p)$

Vedclass · Answer

(A) Let the angle between the lines represented by $x^2 - 2pxy + y^2 = 0$ be $	heta$.Comparing the given equation with $ax^2 + 2hxy + by^2 = 0$,we get $a = 1, b = 1$,and $2h = -2p$,which implies $h = -p$.The formula for the angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values,we get $	an 	heta = \left| \frac{2\sqrt{(-p)^2 - (1)(1)}}{1 + 1} ight| = \left| \frac{2\sqrt{p^2 - 1}}{2} ight| = \sqrt{p^2 - 1}$.Since $	an^2 	heta = p^2 - 1$,we have $p^2 = 1 + 	an^2 	heta = \sec^2 	heta$.Thus,$\sec 	heta = p$,which means $	heta = \sec^{-1}(p)$.

Find the angle between the lines represented by the equation $x^2 - 2pxy + y^2 = 0$.

Similar Questions

The angle between the lines represented by the equation ${x^2} - xy - 6{y^2} - 7x + 31y - 18 = 0$ is.....$^o$

Find the angle between the pair of lines represented by the equation $x^2+4xy+y^2=0$. (in $^{\circ}$)

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$,where $\lambda$ is a real number,represents a pair of lines. If $\theta$ is the acute angle between the lines,then $\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}} = $

If the pair of lines given by $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ are perpendicular to each other,then $\alpha$ is

If $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$,then the angle between $x^2+2 x y \sec \theta+y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module