The slope of one of the pair of lines 2x^2 + | vedclass

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

Question

(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Let the slopes of the two lines be $m$ and $3m$. The equation of the pair of l

Vedclass · Accepted Answer

$\pm 8$

Vedclass · Answer

(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Let the slopes of the two lines be $m$ and $3m$. The equation of the pair of lines can be written as $(y - mx)(y - 3mx) = 0$. Expanding this,we get $y^2 - 4mxy + 3m^2x^2 = 0$,or $3m^2x^2 - 4mxy + y^2 = 0$. Dividing the original equation $2x^2 + hxy + 6y^2 = 0$ by $6$,we get $\frac{1}{3}x^2 + \frac{h}{6}xy + y^2 = 0$. Comparing the coefficients of $x^2$ and $xy$ with $3m^2x^2 - 4mxy + y^2 = 0$,we have: $3m^2 = \frac{1}{3}$ $\Rightarrow m^2 = \frac{1}{9}$ $\Rightarrow m = \pm \frac{1}{3}$. Also,$-4m = \frac{h}{6} \Rightarrow h = -24m$. Substituting $m = \pm \frac{1}{3}$ into the expression for $h$: $h = -24(\pm \frac{1}{3}) = \mp 8$. Thus,$h = \pm 8$.

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

Similar Questions

The joint equation of the straight lines $x + y = 1$ and $x - y = 4$ is

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is

If the slope of one of the lines represented by $ax^2+2hxy+by^2=0$ is twice that of the other,then $h^2:ab$ is

If $s$ and $p$ are respectively the sum and the product of the slopes of the lines $3x^2 - 2xy - 15y^2 = 0$,then $s:p$ is equal to

The equation $r^2 \cos^2 \left(\theta - \frac{\pi}{3}\right) = 2$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module