If the tangent to the parabola y2 = x at a point
(\[\alpha \], \[\beta \]), (\[\beta \] > 0) is also a tangent to the ellipse,
x2 + 2y2 = 1, then \[\alpha \] is equal to :
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Official solution
Point P(\[\alpha \], \[\beta \]) is on the parabola y2 = x
\[ \therefore \] \[{\beta ^2} = \alpha \] ...........(1)
Equation of tangent to the parabola y2 = x
at (\[\alpha \], \[\beta \]) is T = 0
\[\beta y = {{x + \alpha } \over 2}\]
\[ \Rightarrow \] \[2\beta y = x + \alpha \]
\[ \Rightarrow \] \[y = {1 \over {2\beta }}x + {\alpha \over {2\beta }}\]
For this straight line, m = \[{1 \over {2\beta }}\] and c = \[{\alpha \over {2\beta }}\]
This tangent is also a tangent to ellipse x2 + 2y2 = 1.
\[ \Rightarrow \] \[{{{x^2}} \over 1} + {{{y^2}} \over {{1 \over 2}}} = 1\]
\[ \therefore \] a2=1 and \[{b^2} = {1 \over 2}\].
We know the condition of tangency on ellipse is
c2=a2m2+b2
\[ \Rightarrow \] \[{\left( {{\alpha \over {2\beta }}} \right)^2} = 1{\left( {{1 \over {2\beta }}} \right)^2} + {1 \over 2}\]
\[ \Rightarrow \] \[{{{\alpha ^2}} \over {4{\beta ^2}}} = {1 \over {4{\beta ^2}}} + {1 \over 2}\]
\[ \Rightarrow \] \[{\alpha ^2} = 1 + 2{\beta ^2}\]
\[ \Rightarrow \] \[{\alpha ^2} = 1 + 2\alpha \]
\[ \Rightarrow \] \[{\alpha ^2} - 2\alpha - 1 = 0\]
\[ \Rightarrow \] \[\alpha = 1 + \sqrt 2 \] and \[\alpha = 1 - \sqrt 2 \]