# Locus of mid point of the portion between the axes of $\dpi{100} x\; \cos \alpha +y\; \sin \alpha =p\; where\; p$  is constant is Option 1) $x^{2}+y^{2}=\frac{4}{p^{2}}\;$ Option 2) $\; x^{2}+y^{2}=4p^{2}\;$ Option 3) $\; \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{2}{p^{2}}\;$ Option 4) $\; \; \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}}$

As we learnt in

Normal form -

$xcos\omega + ysin\omega = p$

- wherein

p is the length of perpendicular segment from origin and $\omega$ is the angle made by this perpendicular with +ve $x$-axis.

$x\cos \alpha +y\sin \alpha =p$

Point A is:  $x=p\sec \alpha , \: y=0$

Point B is: $x=0, \: y=p cosec\alpha$

We have, $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{p^{2}}$

Option 1)

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

This option is incorrect

Option 2)

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

This option is incorrect

Option 3)

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

This option is incorrect

Option 4)

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

This option is correct

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