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If the line $p x+q y=0$ coincides with one of the lines given by $a x^{2}+2 h x y+b y^{2}=0$, then
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Verified Answer
The correct answer is:
$a q^{2}-2 h p q+b p^{2}=0$
The equation $a x^{2}+2 h x y+b y^{2}=0$ can be rewritten as
$$
a+\frac{2 h y}{x}+b\left(\frac{y}{x}\right)^{2}=0
$$
Consider,
$$
p x+q y=0
$$
$$
p x=-q y \Rightarrow \frac{y}{x}=\frac{-p}{q}
$$
Substituting in Eq. (i), we get
$$
\begin{aligned}
& a+2 h\left(-\frac{p}{q}\right)+b\left(\frac{-p}{q}\right)^{2}=0 \Rightarrow a-\frac{2 h p}{q}+b \frac{p^{2}}{q^{2}}=0 \\
\Rightarrow & a q^{2}-2 h p q+b p^{2}=0
\end{aligned}
$$
$$
a+\frac{2 h y}{x}+b\left(\frac{y}{x}\right)^{2}=0
$$
Consider,
$$
p x+q y=0
$$
$$
p x=-q y \Rightarrow \frac{y}{x}=\frac{-p}{q}
$$
Substituting in Eq. (i), we get
$$
\begin{aligned}
& a+2 h\left(-\frac{p}{q}\right)+b\left(\frac{-p}{q}\right)^{2}=0 \Rightarrow a-\frac{2 h p}{q}+b \frac{p^{2}}{q^{2}}=0 \\
\Rightarrow & a q^{2}-2 h p q+b p^{2}=0
\end{aligned}
$$
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