Respuesta :
The equation of the pair of lines perpendicular to the lines given equation is [tex]b x^{2}-2 h x y+a y^{2}=0[/tex].
Solution:
Given equation is [tex]a x^{2}+2 h x y+b y^{2}=0[/tex].
Let [tex]m_1[/tex] and [tex]m_2[/tex] be the slopes of the given lines.
Sum of the roots = [tex]-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}[/tex]
[tex]$m_1+m_2=\frac{-2h}{b}[/tex] – – – – – (1)
Product of the roots = [tex]-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}[/tex]
[tex]$m_1 \cdot m_2=\frac{a}{b}[/tex] – – – – – (2)
The required lines are perpendicular to these lines.
Slopes of the required lines are [tex]$-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}[/tex]
Required lines also passes through the origin,
therefore their y-intercepts are 0.
Hence their equations are:
[tex]$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x[/tex]
Do cross multiplication, we get
[tex]m_1y=-x \ \text{and} \ m_2y=-x[/tex]
Add x on both sides of the equation, we get
[tex]x+m_1y=0 \ \text{and} \ x+m_2y=0[/tex]
Therefore, the joint equation of the line is
[tex]\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0[/tex]
[tex]x^2+m_2xy+m_1xy+m_1m_2y^2=0[/tex]
[tex]x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0[/tex]
Substitute (1) and (2), we get
[tex]$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0[/tex]
To make the denominator same, multiply and divide first term by b.
[tex]$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0[/tex]
[tex]$\frac{bx^2-2hxy+ay^2}{b} = 0[/tex]
Do cross multiplication, we get
[tex]b x^{2}-2 h x y+a y^{2}=0[/tex]
Hence equation of the pair of lines perpendicular to the lines given equation is [tex]b x^{2}-2 h x y+a y^{2}=0[/tex].
