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Locus of mid point of the portion between the axes of 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}}

 

Answers (1)

best_answer

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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