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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    A tangent to the circle \[{{x}^{2}}+{{y}^{2}}=4\]intersects the hyperbola \[{{x}^{2}}-2{{y}^{2}}=2\]at P and Q. If locus of mid- point of PQ is \[{{({{x}^{2}}-2{{y}^{2}})}^{2}}=\lambda ({{x}^{2}}+4{{y}^{2}}),\]then \[\lambda \] equals

    A)  4                                

    B)  2

    C)  \[\frac{1}{2}\]                                     

    D)  \[\frac{1}{4}\]

    Correct Answer: A

    Solution :

    Equation of chord of hyperbola \[\frac{{{x}^{2}}}{2}\,-\frac{{{y}^{2}}}{1}=1,\]whose mid-point is (h, k) is          \[\frac{hx}{2}-ky=\frac{{{h}^{2}}}{2}\,-\frac{{{k}^{2}}}{1}\] (using \[T={{S}_{1}}\] ) As it is tangent to the circle \[{{x}^{2}}+{{y}^{2}}=4,\] so \[\left| \frac{\frac{{{h}^{2}}}{2}-{{k}^{2}}}{\sqrt{\frac{{{h}^{2}}}{4}+{{k}^{2}}}} \right|\,=2\Rightarrow \,{{\left( \frac{{{h}^{2}}}{2-{{k}^{2}}} \right)}^{2}}\,=4\left( \frac{{{h}^{2}}}{4}\,+{{k}^{2}} \right)\] \[\Rightarrow \] Locus of (h, k) is \[{{({{x}^{2}}\,-2{{y}^{2}})}^{2}}=4({{x}^{2}}+4{{y}^{2}})\] \[\therefore \,\,\lambda =4\]


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