JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    The equation of a tangent to the parabola, \[{{x}^{2}}=8y,\]which makes an angle \[\theta \] with the positive direction of x-axis is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[x=y\cot \theta -2\tan \theta \]               

    B) \[y=x\tan \theta -2\cot \theta \]

    C) \[x=y\cot \theta +2\tan \theta \]

    D)               \[y=x\tan \theta +2\cot \theta \]

    Correct Answer: C

    Solution :

    Given equation of parabola is \[{{x}^{2}}=8y\] \[\therefore \]\[\frac{8dy}{dx}=2x\Rightarrow \frac{dy}{dx}=\frac{x}{4}=\tan \theta \](Given) \[\Rightarrow x=4\tan \theta \]\[\therefore \]\[y=2{{\tan }^{2}}\theta \] Now, equation of tangent at \[(4\tan \theta ,2{{\tan }^{2}}\theta )\]is\[y-2{{\tan }^{2}}\theta =\tan \theta \]\[(x-4\tan \theta )\] \[\Rightarrow \]\[x=y\cot \theta +2\tan \theta \]


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