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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$. When the axes are rotated by an angle $	heta$,the new coordinates $(x', y')$ are related to the old coordinates $(x, y)$ by $x = x' \cos 	heta - y' \sin 	heta$ and $y = x' \sin 	heta + y' \cos 	heta$. Substituting these into the equation of the line: $\frac{x' \cos 	heta - y' \sin 	heta}{a} + \frac{x' \sin 	heta + y' \cos 	heta}{b} = 1$ $x' \left( \frac{\cos 	heta}{a} + \frac{\sin 	heta}{b} ight) + y' \left( \frac{\cos 	heta}{b} - \frac{\sin 	heta}{a} ight) = 1$. Comparing this with the intercept form of the line in the new axes,$\frac{x'}{p} + \frac{y'}{q} = 1$,we get: $\frac{1}{p} = \frac{\cos 	heta}{a} + \frac{\sin 	heta}{b}$ and $\frac{1}{q} = \frac{\cos 	heta}{b} - \frac{\sin 	heta}{a}$. Squaring and adding these equations: $\frac{1}{p^2} + \frac{1}{q^2} = \left( \frac{\cos 	heta}{a} + \frac{\sin 	heta}{b} ight)^2 + \left( \frac{\cos 	heta}{b} - \frac{\sin 	heta}{a} ight)^2$ $= \frac{\cos^2 	heta}{a^2} + \frac{\sin^2 	heta}{b^2} + \frac{2 \sin 	heta \cos 	heta}{ab} + \frac{\cos^2 	heta}{b^2} + \frac{\sin^2 	heta}{a^2} - \frac{2 \sin 	heta \cos 	heta}{ab}$ $= \frac{\cos^2 	heta + \sin^2 	heta}{a^2} + \frac{\cos^2 	heta + \sin^2 	heta}{b^2}$ $= \frac{1}{a^2} + \frac{1}{b^2}$. Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

$A$ line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle $\theta$ keeping the origin fixed,this line $L$ has the intercepts $p$ and $q$. Then

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