The locus of the midpoint of the portion of t | vedclass

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(B) Given the line equation: $x \cos \alpha + y \sin \alpha = p$ ...$(i)$Let $P(h, k)$ be the midpoint of the portion of the line intercepted by the c

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$\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$

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(B) Given the line equation: $x \cos \alpha + y \sin \alpha = p$ ...$(i)$Let $P(h, k)$ be the midpoint of the portion of the line intercepted by the coordinate axes.When $x = 0$,the line meets the $y$-axis at $y = \frac{p}{\sin \alpha}$. So,point $B$ is $(0, \frac{p}{\sin \alpha})$.When $y = 0$,the line meets the $x$-axis at $x = \frac{p}{\cos \alpha}$. So,point $A$ is $(\frac{p}{\cos \alpha}, 0)$.The midpoint $P(h, k)$ is given by $(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha})$.Thus,$h = \frac{p}{2 \cos \alpha} \Rightarrow \cos \alpha = \frac{p}{2h}$ and $k = \frac{p}{2 \sin \alpha} \Rightarrow \sin \alpha = \frac{p}{2k}$.Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get:$(\frac{p}{2h})^2 + (\frac{p}{2k})^2 = 1$$\frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1$$\frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

The locus of the midpoint of the portion of the line $x \cos \alpha + y \sin \alpha = p$ intercepted by the coordinate axes,where $p$ is a constant,is

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