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(B) The equation of the line $L$ with respect to the original axes is $\frac{x}{a} + \frac{y}{b} = 1$.When the axes are rotated by an angle $\alpha$,t

Vedclass · Accepted Answer

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

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(B) The equation of the line $L$ with respect to the original axes 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 terms:$x'\left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right) + y'\left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right) = 1$Comparing this to the intercept form $\frac{x'}{p} + \frac{y'}{q} = 1$,we get:$\frac{1}{p} = \frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}$ 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} = \left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right)^2 + \left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right)^2$$= \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha}{b^2} + \frac{2\sin \alpha \cos \alpha}{ab} + \frac{\cos^2 \alpha}{b^2} + \frac{\sin^2 \alpha}{a^2} - \frac{2\sin \alpha \cos \alpha}{ab}$$= \frac{\cos^2 \alpha + \sin^2 \alpha}{a^2} + \frac{\sin^2 \alpha + \cos^2 \alpha}{b^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle keeping the origin fixed,the same line $L$ has intercepts $p$ and $q$. Then:

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