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JEE Main & Advanced Mathematics Straight Line Question Bank Distance between two lines, Perpendicular distance of the line from a point Position of point w.r.t. line

  • question_answer
    If p and \[p'\]be the distances of origin from the lines \[x\sec \alpha +y\text{cosec }\alpha =k\] and \[x\cos \alpha -y\sin \alpha =k\cos 2\alpha \], then \[4{{p}^{2}}+{{{p}'}^{2}}\]= 

    A)            k     

    B)            \[2k\]

    C)            \[{{k}^{2}}\]                         

    D)            \[2{{k}^{2}}\]

    Correct Answer: C

    Solution :

               Here\[p=\left| \frac{-k}{\sqrt{{{\sec }^{2}}\alpha +\text{cose}{{\text{c}}^{2}}\alpha }} \right|\], \[{p}'=\left| \frac{-k\cos 2\alpha }{\sqrt{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }} \right|\]                    Hence \[4{{p}^{2}}+{{{p}'}^{2}}=\frac{4{{k}^{2}}}{{{\sec }^{2}}\alpha +\text{cose}{{\text{c}}^{2}}\alpha }+\frac{{{k}^{2}}{{({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )}^{2}}}{1}\]                    \[=4{{k}^{2}}{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +{{k}^{2}}({{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha )\]\[-2{{k}^{2}}{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \]                    \[={{k}^{2}}{{({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}^{2}}={{k}^{2}}\].


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