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

  • question_answer
    If the tangent drawn at a point \[({{t}^{2}},2t)\]on the parabola \[{{y}^{2}}=4x\] is same as normal drawn at \[(\sqrt{5}cos\alpha ,2sin\alpha )\]on the ellipse\[\frac{{{x}^{2}}}{5}+\frac{{{y}^{2}}}{4}=1\], then which of the following is not true?

    A) \[t=\pm \frac{1}{\sqrt{5}}\]       

    B) \[\alpha =-{{\tan }^{-1}}2\]

    C) \[\alpha ={{\tan }^{-1}}2\]      

    D) None of these

    Correct Answer: D

    Solution :

    [d]: Equation of tangent at \[({{t}^{2}},2t)\]is \[x=ty-{{t}^{2}}\]                        ...(i) Equation of normal to ellipse at \[(\sqrt{5}cos\alpha ,2sin\alpha )\] is \[\sqrt{5}x\sec \alpha -2y\operatorname{cosec}\alpha =1\]                      ...(ii) Now, (i) = (ii) \[\Rightarrow \]\[\sqrt{5}\sec \alpha =\frac{2\cos ec\alpha }{t}=-\frac{1}{{{t}^{2}}}\] \[\Rightarrow \]\[\cos \alpha =-\sqrt{5}{{t}^{2}}\]and\[\sin \alpha =-2t\] \[\Rightarrow \]\[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =5{{t}^{4}}+4{{t}^{2}}=1\]\[\Rightarrow \]\[{{t}^{2}}=\frac{1}{5}\] Also,\[\frac{\sin \alpha }{\cos \alpha }=\frac{-2t}{-\sqrt{5}{{t}^{2}}}=\frac{2}{\sqrt{5}}\times \frac{1}{t}=\frac{2}{\sqrt{5}}\times (\pm \sqrt{5})\] \[\therefore \]\[\tan \alpha =\pm 2\Rightarrow \tan \alpha =-2,\tan \alpha =2\]

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