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

If $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel lines,then $\sqrt{\frac{g^2-ac}{f^2-bc}}$ is equal to

The angle between the lines represented by $x^2 + xy = 0$ is .....$^o$

The equation $4x^2 + 12xy + 9y^2 + 2gx + 2fy + c = 0$ represents two real parallel straight lines,if

For $\alpha \in [0, \frac{\pi}{2}]$,the angle between the lines represented by $[x \cos \theta - y][(\cos \theta + \tan \alpha) x - (1 - \cos \theta \tan \alpha) y] = 0$ is

Find the value of $k$,if the angle between the straight lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1}(k)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module