If the slope of one of the two lines represen | vedclass

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

Question

(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Dividing by $x^2$,we get $\frac{1}{b}(\frac{y}{x})^2 + \frac{2}{h}(\fra

Vedclass · Accepted Answer

$9:8$

Vedclass · Answer

(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Dividing by $x^2$,we get $\frac{1}{b}(\frac{y}{x})^2 + \frac{2}{h}(\frac{y}{x}) + \frac{1}{a} = 0$. Let the slopes be $m$ and $2m$. From the sum of roots,$m + 2m = 3m = -\frac{2/h}{1/b} = -\frac{2b}{h}$,so $m = -\frac{2b}{3h}$. From the product of roots,$m \cdot 2m = 2m^2 = \frac{1/a}{1/b} = \frac{b}{a}$. Substituting $m$ in the product equation: $2(-\frac{2b}{3h})^2 = \frac{b}{a}$. $2 \cdot \frac{4b^2}{9h^2} = \frac{b}{a}$. $\frac{8b^2}{9h^2} = \frac{b}{a}$. $\frac{ab}{h^2} = \frac{9}{8}$. Thus,$ab : h^2 = 9:8$.

If the slope of one of the two lines represented by $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$ is twice that of the other,then $ab : h^2 = $

Similar Questions

The equation ${y^2} - {x^2} + 2x - 1 = 0$ represents

If the slope of one of the lines given by $ax^2+2hxy+by^2=0$ is two times the other,then

Let $PQR$ be a right-angled isosceles triangle,right-angled at $P(2, 1)$. If the equation of the line $QR$ is $2x + y = 3$,then the equation representing the pair of lines $PQ$ and $PR$ is

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

If the slopes of the lines given by the equation $ax^{2} + 2hxy + by^{2} = 0$ are in the ratio $5:3$,then the ratio $h^{2}:ab$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module