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(D) The equation of the line $L$ with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$. When the axes are rotated by an angle $\alpha$,the co

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$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$

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(D) The equation of the line $L$ with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$. When the axes are rotated by an angle $\alpha$,the coordinates $(x, y)$ transform to $(x', y')$ where $x = x' \cos \alpha - y' \sin \alpha$ and $y = x' \sin \alpha + y' \cos \alpha$. Substituting these into the line equation: $\frac{x' \cos \alpha - y' \sin \alpha}{a} + \frac{x' \sin \alpha + y' \cos \alpha}{b} = 1$. Rearranging gives $x'(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}) + y'(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}) = 1$. Since the new intercepts are $p$ and $q$,we have $\frac{1}{p} = \frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}$ and $\frac{1}{q} = \frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}$. Squaring and adding these equations: $\frac{1}{p^2} + \frac{1}{q^2} = (\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b})^2 + (\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a})^2$. Expanding the terms: $\frac{1}{p^2} + \frac{1}{q^2} = \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha}{b^2} + \frac{2 \sin \alpha \cos \alpha}{ab} + \frac{\cos^2 \alpha}{b^2} + \frac{\sin^2 \alpha}{a^2} - \frac{2 \sin \alpha \cos \alpha}{ab}$. Simplifying: $\frac{1}{p^2} + \frac{1}{q^2} = \frac{1}{a^2}(\cos^2 \alpha + \sin^2 \alpha) + \frac{1}{b^2}(\sin^2 \alpha + \cos^2 \alpha) = \frac{1}{a^2} + \frac{1}{b^2}$.

Column-$I$	Column-$II$
$A$. Christmas tree	$i$. Bennettites
$B$. Fossil plant	$ii$. Cycas
$C$. Ornamental plant	$iii$. Araucaria
$D$. Tallest tree	$iv$. Sequoia
	$v$. Thuja

$A$ line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the coordinate axes are rotated through an angle $\alpha$ keeping the origin fixed,the same line $L$ has intercepts $p$ and $q$ on the new axes. Then

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