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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    A tangent to ellipse \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\] at any point P meets the line \[x=0\] at point Q. Let R be the image of Q in the line \[y=x,\] then circle whose extremities of diameter are Q and R, passes through a fixed point, whose coordinate is

    A)  (3, 0)                                    

    B)  (5, 0)

    C)   (0, 0)                                   

    D)  (4, 0)

    Correct Answer: C

    Solution :

    The equation of tangent at P \[(5\cos \theta ,4\sin \theta )\]is \[\frac{x}{5}\cos \theta +\frac{y}{4}\sin \theta =1\].Point Q is \[(0,4\,\cos ec\theta )\]and R is \[(4\,\cos ec\theta ,0)\]equation of circle is \[x(x-4\,\cos ec\theta )+y(y-4\cos ec\,\theta )=0\] i.e.\[{{x}^{2}}+{{y}^{2}}-4(x+y)\cos ec\theta =0\]


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