question_answer

If the tangent to the parabola \[{{y}^{2}}=x\]at a point \[(\alpha ,\beta ),(\beta >0)\]is also a tangent to the ellipse, \[{{x}^{2}}+2{{y}^{2}}=1,\] then \[\alpha \]is equal to :                                                                         [JEE Main 9-4-2019 Afternoon]

A) \[2\sqrt{2}+1\]

B)               \[\sqrt{2}-1\]

C) \[\sqrt{2}+1\]

D) \[2\sqrt{2}-1\]

Correct Answer: C

Solution :

\[T:y(\beta )=\frac{1}{2}(x+{{\beta }^{2}})\]           \[2y\beta =x+{{\beta }^{2}}\]           \[y=\left( \frac{1}{2\beta } \right)x+\frac{\beta }{2}\]           \[m=\frac{1}{2\beta };C=\frac{\beta }{2}\]           \[\frac{\beta }{2}=\pm \sqrt{\frac{1}{2{{\beta }^{2}}}+\frac{1}{2}}\]           \[\frac{{{\beta }^{2}}}{4}=\frac{1}{4{{\beta }^{2}}}+\frac{1}{2}\]           \[\frac{{{\beta }^{2}}}{4}=\frac{1+2{{\beta }^{2}}}{4{{\beta }^{2}}}\]           \[\Rightarrow \]\[{{\beta }^{4}}-2{{\beta }^{2}}-1=0\]           \[{{({{\beta }^{2}}-1)}^{2}}=2\]           \[{{\beta }^{2}}-1=\sqrt{2}\]           \[{{\beta }^{2}}=\sqrt{2}+1\]