For l in R,the equation (2 l-3) x^2+2 l x y-y | vedclass

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

Question

(B) The general second-degree equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines if $h^2 - ab \geq 0$.Comparing the given equation

Vedclass · Accepted Answer

for all values of $l \in R-(-3,1)$

Vedclass · Answer

(B) The general second-degree equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines if $h^2 - ab \geq 0$.Comparing the given equation $(2l-3)x^2 + 2lxy - y^2 = 0$ with the standard form,we have $a = (2l-3)$,$h = l$,and $b = -1$.The condition for representing a pair of lines is $h^2 - ab \geq 0$.Substituting the values,we get $l^2 - (2l-3)(-1) \geq 0$.$l^2 + (2l-3) \geq 0$$l^2 + 2l - 3 \geq 0$$(l+3)(l-1) \geq 0$.Solving this inequality,we find that the expression is non-negative when $l \in (-\infty, -3] \cup [1, \infty)$,which can be written as $l \in R - (-3, 1)$.

For $l \in R$,the equation $(2 l-3) x^2+2 l x y-y^2=0$ represents a pair of lines

Similar Questions

If the slope of one line of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line,then $h =$

The joint equation of the lines through the origin trisecting the angles in the first and third quadrants is

The equation of the pair of lines perpendicular to the lines represented by $x^2-5xy+4y^2=0$ and passing through the point $(2,1)$ is

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

The equation of a pair of lines is given by a second-degree homogeneous equation. If one of them is perpendicular to the line $x+2y+7=0$ and the other is parallel to the line $3x+4y+5=0$,then the equation of that pair of lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module