The angle between the lines represented by th | vedclass

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

Question

(A) The given equation is $x^2 - 2pxy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$h = -p$,and $b = 1$.L

Vedclass · Accepted Answer

$\sec^{-1} p$

Vedclass · Answer

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

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

Similar Questions

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

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

If the pair of straight lines $6x^2 - 5xy + y^2 = 0$ makes angles $\alpha$ and $\beta$ with the $X$-axis,then $\tan(\alpha - \beta) = $

The pair of straight lines represented by the equation $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$ are

The lines $ax^2+2hxy+by^2=0$ are at right angles if

Vedclass Test Series

Exam Paper Generator

Online Exam Module