The pair of straight lines is represented by | vedclass

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

Question

(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. For a general equation of a pair of lines $Ax^2 + 2Hxy + By^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. For a general equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$,the lines are perpendicular if $A + B = 0$. Here,$A = 3d$ and $B = d^2 - 2$. Setting $A + B = 0$,we get $3d + d^2 - 2 = 0$,which is $d^2 + 3d - 2 = 0$. Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we find $d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}$. Since there are two distinct real values for $d$,the condition is satisfied for $2$ values of $d$. Thus,option $B$ is correct.

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?

Similar Questions

The angle between the pair of straight lines $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

The absolute value of the tangent of the difference of the angles made by the lines $4x^2 - 24xy + 11y^2 = 0$ with the $X$-axis is

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

If the angle between the lines represented by the equation $y^2 + kxy - x^2 \tan^2 A = 0$ is $2A$,then $k = $

If $ax^2+6xy+by^2-10x+10y-6=0$ represents a pair of perpendicular lines,then the value of $|a|$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module