If the slope of one of the lines represented | vedclass

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

Question

(A) The equation of the pair of lines is $2x^2 + 3xy + ky^2 = 0$. Dividing by $x^2$,we get $k(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$. Let $m = \frac

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) The equation of the pair of lines is $2x^2 + 3xy + ky^2 = 0$. Dividing by $x^2$,we get $k(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$. Let $m = \frac{y}{x}$ be the slope of the lines. Then $km^2 + 3m + 2 = 0$. Given that one slope $m_1 = 2$,substituting this into the equation: $k(2)^2 + 3(2) + 2 = 0 \implies 4k + 6 + 2 = 0 \implies 4k = -8 \implies k = -2$. The equation becomes $2x^2 + 3xy - 2y^2 = 0$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 2$,$2h = 3 \implies h = \frac{3}{2}$,and $b = -2$. The angle $	heta$ between the lines is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$. Here $a + b = 2 + (-2) = 0$. Since the sum of coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular. Therefore,$	heta = \frac{\pi}{2}$.

If the slope of one of the lines represented by $2x^2 + 3xy + ky^2 = 0$ is $2$,then the angle between the pair of lines is

Similar Questions

The equation $3ax^2 - 16xy - (a^2 - 10)y^2 = 0$ represents

The acute angle between the lines represented by $({x^2} + {y^2})\sqrt{3} = 4xy$ is

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 $2x^2 + 3xy + y^2 = 0$ makes angles $\theta_1$ and $\theta_2$ with the positive direction of the $X$-axis,then $|\tan(\theta_1 - \theta_2)| = $

The pair of straight lines is represented by the equation $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. If the lines are perpendicular to each other,for how many values of $d$ will this condition be satisfied?

Vedclass Test Series

Exam Paper Generator

Online Exam Module