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

Option: 1

\mathrm{t= \pm \frac{1}{\sqrt{5}}}


Option: 2

\mathrm{\alpha=-\tan ^{-1} 2}


Option: 3

\mathrm{\alpha=\tan ^{-1} 2}


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\text { Equation of tangent to } y^2=4 x \text { at }\left(t^2, 2 t\right) \text { is }}

\mathrm{x=t y-t^2}        ....[i]

\mathrm{\text { Equation of normal to ellipse } \frac{x^2}{5}+\frac{y^2}{4}=1}

\mathrm{\text { at }(\sqrt{5} \cos \alpha, 2 \sin \alpha) \text { is } \sqrt{5} \sec \alpha x-2 y \operatorname{cosec} \alpha=1 \quad \ldots \text { (ii) }}

Given that (i) and (ii) are identical

\mathrm{\begin{aligned} & \therefore \quad \sqrt{5} \sec \alpha=\frac{2 \operatorname{cosec} \alpha}{t}=-\frac{1}{t^2} \\ & \Rightarrow \cos \alpha=-\sqrt{5} t^2 \text { and } \sin \alpha=-2 t \\ & \Rightarrow \cos ^2 \alpha+\sin ^2 \alpha=5 t^4+4 t^2=1 \\ & \Rightarrow t^2=\frac{1}{5} \Rightarrow t= \pm \frac{1}{\sqrt{5}} \therefore \text { (a) is true } \end{aligned}}

\mathrm{\begin{aligned} & \text { and } \frac{\sin \alpha}{\cos \alpha}=-\frac{2 t}{-\sqrt{5} t^2}=\frac{2}{\sqrt{5}} \times \frac{1}{t}=\frac{2}{\sqrt{5}} \times( \pm \sqrt{5}) \\ & \Rightarrow \tan \alpha= \pm 2 \end{aligned}}

Posted by

Kuldeep Maurya

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