If the lines represented by the equation 2x^2 | vedclass

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

Question

(D) The given equation is $2x^2 - 3xy + y^2 = 0$.Dividing by $x^2$,we get $2 - 3(\frac{y}{x}) + (\frac{y}{x})^2 = 0$.Let $m = \frac{y}{x} = 	an 	het

Vedclass · Accepted Answer

$5/4$

Vedclass · Answer

(D) The given equation is $2x^2 - 3xy + y^2 = 0$.Dividing by $x^2$,we get $2 - 3(\frac{y}{x}) + (\frac{y}{x})^2 = 0$.Let $m = \frac{y}{x} = 	an 	heta$. Then $m^2 - 3m + 2 = 0$.The slopes of the lines are $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.From the quadratic equation,the sum of roots $m_1 + m_2 = 3$ and the product of roots $m_1 m_2 = 2$.We need to find $\cot^2 \alpha + \cot^2 \beta = \frac{1}{m_1^2} + \frac{1}{m_2^2} = \frac{m_1^2 + m_2^2}{(m_1 m_2)^2}$.Using the identity $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2$,we get:$\frac{(m_1 + m_2)^2 - 2m_1 m_2}{(m_1 m_2)^2} = \frac{(3)^2 - 2(2)}{(2)^2} = \frac{9 - 4}{4} = \frac{5}{4}$.

If the lines represented by the equation $2x^2 - 3xy + y^2 = 0$ make angles $\alpha$ and $\beta$ with the $x$-axis,then $\cot^2 \alpha + \cot^2 \beta = $

Similar Questions

The joint equation of the lines passing through the origin and trisecting the first quadrant is

The equation $\frac{x^2}{a} + \frac{xy}{h} + \frac{y^2}{b} = 0$ $(a \neq 0, h \neq 0, b \neq 0)$ represents two coincident lines if:

If $4ab = 3h^2$,then the ratio of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

The locus of a point which moves such that its distance from the $x$-axis is twice its distance from the line $x-y=0$ is

The combined equation of the lines passing through the origin making an acute angle $\alpha$ with the line $y=x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module