If the equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents two straight lines equidistant from the…
If the equation
$a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents two straight lines equidistant from the origin, then $f^4-g^4=$
- $b f^2-a g^2$
- $a g^2-b f^2$
- $c\left(b f^2-a g^2\right)$
- $c\left(a f^2-b g^2\right)$
Solution
Let the lines represented by equation
$
\begin{aligned}
& a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\
& \text { are } y=m_1 x+c_1 \text { and } y=m_2 x+c_2 \\
& \begin{aligned}
& \text { So, } a x^2+2 h x y+b y^2+2 g x+2 f y+c \\
&=\left(m_1 x-y+c_1\right)\left(m_1 x-y+c_2\right) \\
& \Rightarrow \quad \frac{m_1 m_2}{a}=\frac{-\left(m_1+m_2\right)}{2 h}=\frac{1}{b} \\
&= \frac{m_1 C_2+m_2 C_1}{2 g}=\frac{-\left(C_1+C_2\right)}{2 f}=\frac{C_1 C_2}{c}
\end{aligned}
\end{aligned}
$
Now, according to the information given in question, as lines are equidistance from origin, so
$
\begin{aligned}
& \frac{\left|C_1\right|}{\sqrt{1+m_1^2}}=\frac{\left|C_2\right|}{\sqrt{1+m_2^2}} \\
& \Rightarrow \quad c_1^2\left(1+m_2^2\right)=c_2^2\left(1+m_1^2\right) \\
& \Rightarrow \quad c_1^2-c_2^2=c_2^2 m_1^2-c_1^2 m_2^2 \\
& \Rightarrow\left(C_1+C_2\right)\left(C_1-C_2\right)=\left(C_2 m_1+C_1 m_2\right) \\
& \left(C_2 m_1-C_1 m_2\right) \\
&
\end{aligned}
$
From Eq. (i), we have
$
C_1+C_2=-\frac{2 f}{b} \text { and } m_1 C_2+m_2 C_1=\frac{2 g}{b}
$
and $C_1 C_2=\frac{c}{b}$ and $m_1 m_2=\frac{9}{b}$
$
\begin{aligned}
& \therefore\left|C_1-C_2\right|=\sqrt{\frac{4 f^2}{b^2}-\frac{4 c}{b}}=2 \sqrt{\frac{f^2-b c}{b^2}}=\frac{2}{|b|} \sqrt{f^2-b c} \\
& \text { and }\left|C_2 m_1-C_1 m_2\right|=\sqrt{\frac{4 g^2}{b^2}-4 \frac{a c}{b^2}}=\frac{2}{|b|} \sqrt{g^2-a c}
\end{aligned}
$
Therefore from Eq. (ii), we have
$
\begin{aligned}
& -\frac{2 f}{b} \times \frac{2}{|b|} \sqrt{f^2-b c}=\frac{2 g}{b} \times \frac{2}{|b|} \sqrt{g^2-a c} \\
\Rightarrow & f^2\left(f^2-b c\right)=g^2\left(g^2-a c\right) \text { (on squaring both sides) } \\
\Rightarrow & f^4-g^4=c\left(b f^2-a g^2\right)
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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