If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :
A
√2+1
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B
√2−1
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C
2√2−1
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D
2√2+1
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Solution
The correct option is A√2+1 Equation of tangent to the parabola y2=x at (α,β) is y=(12β)x+β2 Also, (α,β) lies on the parabola y2=x ∴β2=α Equation of ellipse is x21+y21/2=1 ∴a=1,b=1√2 According to condition for [1] to be tangent to the given ellipse, we have (12β)2+12=β4 ⇒β4−2β2−1=0 ⇒α=β2=√2+1