Question

find the equation of the pair of lines perpendicular to the lines pair represented by the equation ax^2-2hxy+by^2=0 and passing through the origin.​

Answer 1

The equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$.

Solution:

Given equation is $a x^{2}+2 h x y+b y^{2}=0$.

Let $m_1$ and $m_2$ be the slopes of the given lines.

Sum of the roots = $-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}$

               $m_1+m_2=\frac{-2h}{b}$ – – – – – (1)

Product of the roots = $-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}$

                    $m_1 \cdot m_2=\frac{a}{b}$ – – – – – (2)

The required lines are perpendicular to these lines.

Slopes of the required lines are $-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}$

Required lines also passes through the origin,

therefore their y-intercepts are 0.

Hence their equations are:

$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x$

Do cross multiplication, we get

$m_1y=-x \ \text{and} \ m_2y=-x$

Add x on both sides of the equation, we get

$x+m_1y=0 \ \text{and} \ x+m_2y=0$

Therefore, the joint equation of the line is

$\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0$

$x^2+m_2xy+m_1xy+m_1m_2y^2=0$

$x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0$

Substitute (1) and (2), we get

$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

To make the denominator same, multiply and divide first term by b.

$\frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

$\frac{bx^2-2hxy+ay^2}{b} = 0$

Do cross multiplication, we get

$b x^{2}-2 h x y+a y^{2}=0$

Hence equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$.