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,\[\frac{\sin \alpha }{\cos \alpha }=\frac{-2t}{-\sqrt{5}{{t}^{2}}}=\frac{2}{\sqrt{5}}\times \frac{1}{t}=\frac{2}{\sqrt{5}}\times (\pm \sqrt{5})\] \[\therefore \]\[\tan \alpha =\pm 2\Rightarrow \tan \alpha =-2,\tan \alpha =2\]
You need to login to perform this action.
You will be redirected in
3 sec
