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If one of the slopes of the pair of lines $\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}=0$ is $\mathrm{n}$ times the other, then
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Verified Answer
The correct answer is:
$4 \mathrm{nh}^{2}=(\mathrm{n}+1)^{2} \mathrm{ab}$
Let $m$ be the slope of the lines $a x^{2}+2 h x y+b y^{2}=0$, then according to question, other slope will be nm.
$\therefore \quad \mathrm{m}+\mathrm{nm}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$
$\Rightarrow \quad \mathrm{nm}^{2}=\frac{\mathrm{a}}{\mathrm{b}}$
$\Rightarrow \quad \mathrm{m}=\pm \sqrt{\frac{\mathrm{a}}{\mathrm{bn}}}$
$\therefore$ From Eq. (i), we get
$$
\pm \sqrt{\frac{a}{b n}}(1+n)=\frac{-2 h}{b}
$$
On squaring both sides, we get
$$
\begin{aligned}
\frac{\mathrm{a}}{\mathrm{bn}}(1+\mathrm{n})^{2} &=\frac{4 \mathrm{~h}^{2}}{\mathrm{~b}^{2}} \\
\Rightarrow \quad 4 \mathrm{~h}^{2} \mathrm{n} &=\mathrm{ab}(1+\mathrm{n})^{2}
\end{aligned}
$$
$\therefore \quad \mathrm{m}+\mathrm{nm}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$
$\Rightarrow \quad \mathrm{nm}^{2}=\frac{\mathrm{a}}{\mathrm{b}}$
$\Rightarrow \quad \mathrm{m}=\pm \sqrt{\frac{\mathrm{a}}{\mathrm{bn}}}$
$\therefore$ From Eq. (i), we get
$$
\pm \sqrt{\frac{a}{b n}}(1+n)=\frac{-2 h}{b}
$$
On squaring both sides, we get
$$
\begin{aligned}
\frac{\mathrm{a}}{\mathrm{bn}}(1+\mathrm{n})^{2} &=\frac{4 \mathrm{~h}^{2}}{\mathrm{~b}^{2}} \\
\Rightarrow \quad 4 \mathrm{~h}^{2} \mathrm{n} &=\mathrm{ab}(1+\mathrm{n})^{2}
\end{aligned}
$$
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