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A line $\mathrm{L}$ has intercepts $\mathrm{a}$ and $\mathrm{b}$ on the coordinate axes. When the coordinate axes are rotated through an angle $\alpha$ keeping the origin fixed, the same line $\mathrm{L}$ has intercepts $\mathrm{p}$ and $q$ on the new axes. Then
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Verified Answer
The correct answer is:
$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$
When axes are rotated through $\alpha$, new co-ordinates $(x \cos \alpha-y \sin \alpha, x \sin \alpha+y \cos \alpha)$
Original equation of $L: \frac{x}{a}+\frac{y}{b}=1$
With shifted coordinates,
$$
\frac{x \cos \alpha-y \sin \alpha}{a}+\frac{x \sin \alpha+y \cos \alpha}{b}=1
$$
Let intercepts are $p$ and $q$
$\therefore \quad(p, 0)$ and $(0, q)$ lie on line
$$
\begin{aligned}
& \frac{1}{p}=\frac{1}{a} \cos \alpha+\frac{1}{b} \sin \alpha \\
& \frac{1}{q}=\frac{1}{b} \cos \alpha+\frac{1}{(-a)} \sin \alpha
\end{aligned}
$$
Squaring both and adding
$$
\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}
$$
Original equation of $L: \frac{x}{a}+\frac{y}{b}=1$
With shifted coordinates,
$$
\frac{x \cos \alpha-y \sin \alpha}{a}+\frac{x \sin \alpha+y \cos \alpha}{b}=1
$$
Let intercepts are $p$ and $q$
$\therefore \quad(p, 0)$ and $(0, q)$ lie on line
$$
\begin{aligned}
& \frac{1}{p}=\frac{1}{a} \cos \alpha+\frac{1}{b} \sin \alpha \\
& \frac{1}{q}=\frac{1}{b} \cos \alpha+\frac{1}{(-a)} \sin \alpha
\end{aligned}
$$
Squaring both and adding
$$
\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}
$$
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