JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer The equation of tangent to \[4{{x}^{2}}-9{{y}^{2}}=36\] which are perpendicular to straight line\[5x+2y-10=0\]are

    A) \[5(y-3)2=\left( x-\frac{\sqrt{117}}{2} \right)\]

    B) \[2y-5x+10-2\sqrt{18}=0\]

    C) \[2y-5x-10-2\sqrt{18}=0\]

    D)  None of these

    Correct Answer: D

    Solution :

     Slope of the equations\[4{{x}^{2}}-9{{y}^{2}}=36\] \[8x-18y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}\]or\[{{m}_{1}}=\frac{4x}{9y}\] Slope of the straight line,\[5x+2y-10=0\]is\[{{m}_{2}}=-\frac{5}{2}\] Therefore, for the perpendicularity,\[{{m}_{1}}{{m}_{2}}=-1\] Now,     \[\frac{4x}{9y}\times \frac{-5}{2}=-1\Rightarrow y=\frac{10x}{9}\] Putting \[y=\frac{10x}{9}\] in \[4{{x}^{2}}-9{{y}^{2}}=36\] gives imaginary roots resulting in no tangents.


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