text: • # 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
  
  [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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