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To the lines $a x^{2}+2 h x y+b y^{2}=0$, the line $a^{2} x^{2}+2 h(a+b) x y+b^{2} y^{2}=0$ are
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The correct answer is:
equally inclined
The equation of the bisectors of the angle between the lines given by $a x^{2}+2 h x y+b y^{2}=0$ is
$$
\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}...(i)
$$
And the equation of the bisectors of the angle between the lines given by $a^{2} x^{2}+2 h(a+b) x y+b^{2} y^{2}=0$ is
$$
\begin{array}{l}
\frac{x^{2}-y^{2}}{a^{2}-b^{2}}=\frac{x y}{h(a+b)} \\
\Rightarrow \frac{x^{2}-y^{2}}{a-b}=\frac{x}{h}...(ii)
\end{array}
$$
From eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined.
$$
\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}...(i)
$$
And the equation of the bisectors of the angle between the lines given by $a^{2} x^{2}+2 h(a+b) x y+b^{2} y^{2}=0$ is
$$
\begin{array}{l}
\frac{x^{2}-y^{2}}{a^{2}-b^{2}}=\frac{x y}{h(a+b)} \\
\Rightarrow \frac{x^{2}-y^{2}}{a-b}=\frac{x}{h}...(ii)
\end{array}
$$
From eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined.
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