JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    The locus of the midpoint of the intercept of the line \[x\cos \alpha +y\sin \alpha =p\] between the coordinate axes is

    A) \[{{x}^{-2}}+{{y}^{-2}}=4{{p}^{-2}}\]

    B) \[{{x}^{-2}}+{{y}^{-2}}={{p}^{-2}}\]

    C) \[{{x}^{2}}+{{y}^{2}}=4{{p}^{-2}}\]

    D) \[{{x}^{2}}+{{y}^{2}}={{p}^{2}}\]

    Correct Answer: A

    Solution :

    [a] : Coordinates of \[A\equiv \left( 0,\frac{p}{\sin \alpha } \right)\] Coordinates of \[B\equiv \left( \frac{p}{\cos \alpha },0 \right)\] Coordinates of \[M\equiv \left( \frac{p}{2\cos \alpha },\frac{p}{2\sin \alpha } \right)\] Now, let \[x=\frac{p}{2\cos \alpha }\]                            ....(i) , \[y=\frac{p}{2\sin \alpha }\]                                           ...(ii)             Since, \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1\]\[\Rightarrow \]\[{{\left( \frac{p}{2x} \right)}^{2}}+{{\left( \frac{p}{2y} \right)}^{2}}=1\] \[\Rightarrow \]\[\frac{{{p}^{2}}}{4}\left[ \frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}{{y}^{2}}} \right]=1\Rightarrow {{x}^{-2}}+{{y}^{-2}}=4{{p}^{-2}}\]

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