The slopes of the lines represented by x^2+2 | vedclass

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

Question

(B) Given the equation of the pair of lines: $x^2+2 h x y+2 y^2=0$ $(i)$.Comparing this with the general equation $a x^2+2 h x y+b y^2=0$,we get $a=1$

Vedclass · Accepted Answer

$\pm \frac{3}{2}$

Vedclass · Answer

(B) Given the equation of the pair of lines: $x^2+2 h x y+2 y^2=0$ $(i)$.Comparing this with the general equation $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$.Let the slopes of the lines be $m_1$ and $m_2$.Then,$m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h$ $(ii)$ and $m_1 m_2 = \frac{a}{b} = \frac{1}{2}$ $(iii)$.Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.Substituting $m_2 = 2m_1$ into $(iii)$,we get $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.If $m_1 = \frac{1}{2}$,then $m_2 = 1$,so $m_1+m_2 = \frac{3}{2}$. From $(ii)$,$-h = \frac{3}{2} \Rightarrow h = -\frac{3}{2}$.If $m_1 = -\frac{1}{2}$,then $m_2 = -1$,so $m_1+m_2 = -\frac{3}{2}$. From $(ii)$,$-h = -\frac{3}{2} \Rightarrow h = \frac{3}{2}$.Thus,$h = \pm \frac{3}{2}$.

The slopes of the lines represented by $x^2+2 h x y+2 y^2=0$ are in the ratio $1:2$,then $h$ is equal to:

Similar Questions

Arrange the following compounds in the increasing order of their dipole moment:
$(I)$ Toluene
$(II)$ $m$-Dichlorobenzene
$(III)$ $o$-Dichlorobenzene
$(IV)$ $p$-Dichlorobenzene

Product $(A)$ is

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$,then the coefficient of $x$ in the cubic equation whose roots are $\alpha(\beta+\gamma), \beta(\gamma+\alpha)$ and $\gamma(\alpha+\beta)$ is

Which of the following is not included in the five-kingdom classification system?

The image of the line $x+y-2=0$ in the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module