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If $p$ and $q$ are the $x$ and $y$-intercepts respectively of the line passing through the points $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$, then $\frac{a^2}{p^2}+\frac{b^2}{q^2}=$
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The correct answer is:
$\sec ^2\left(\frac{\alpha-\beta}{2}\right)$
The equation of line is: $\frac{x}{p}+\frac{y}{q}=1$
Equation (i) passes through
$(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$.
$\therefore \frac{a \cos \alpha}{p}+\frac{b \sin \alpha}{q}=1...(i)$
and $\frac{a \cos \beta}{p}+\frac{b \sin \beta}{q}=1...(ii)$
Solving equations (ii) and (iii), we get:
$\frac{a}{p}=\frac{\sin \beta-\sin \alpha}{\sin (\alpha-\beta)} \text { and } \frac{b}{q}=\frac{\cos \alpha-\cos \beta}{\sin (\alpha-\beta)}$
Now, $\frac{a^2}{p^2}+\frac{b^2}{q^2}=\frac{(\sin \beta-\sin \alpha)^2+(\cos \alpha-\cos \beta)^2}{\sin ^2(\alpha-\beta)^2}$
$\Rightarrow \frac{a^2}{p^2}+\frac{b^2}{q^2}=\sec ^2\left(\frac{\alpha-\beta}{2}\right)$
Equation (i) passes through
$(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$.
$\therefore \frac{a \cos \alpha}{p}+\frac{b \sin \alpha}{q}=1...(i)$
and $\frac{a \cos \beta}{p}+\frac{b \sin \beta}{q}=1...(ii)$
Solving equations (ii) and (iii), we get:
$\frac{a}{p}=\frac{\sin \beta-\sin \alpha}{\sin (\alpha-\beta)} \text { and } \frac{b}{q}=\frac{\cos \alpha-\cos \beta}{\sin (\alpha-\beta)}$
Now, $\frac{a^2}{p^2}+\frac{b^2}{q^2}=\frac{(\sin \beta-\sin \alpha)^2+(\cos \alpha-\cos \beta)^2}{\sin ^2(\alpha-\beta)^2}$
$\Rightarrow \frac{a^2}{p^2}+\frac{b^2}{q^2}=\sec ^2\left(\frac{\alpha-\beta}{2}\right)$
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