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 · 2019 · 09 Apr 2019 · Shift 2 · Parabola

If the tangent to the parabola

y^{2} = x

at a point

(α, β), (β > 0)

is also a tangent to the ellipse,

x^{2} + 2 y^{2} = 1

, then

α

is equal to

(A) $22 + 1$
(B) $2 - 1$
(C) $2 + 1$ ✓ Correct
(D) $22 - 1$

Answer: (C)

Solution

(α, β)

y^{2} = x

the tangent is

y β = \frac{x + α}{2}

; since

β^{2} = α

, this is a line of slope

m = \frac{1}{2 β}

and intercept

c = \frac{β}{2}

. For it to touch the ellipse

x^{2} + 2 y^{2} = 1

, where

a^{2} = 1

and

b^{2} = \frac{1}{2}

, the tangency condition

c^{2} = a^{2} m^{2} + b^{2}

must hold:

\frac{β ^{2}}{4} = \frac{1}{4 β ^{2}} + \frac{1}{2}

. Multiplying through by

4 β^{2}

gives

β^{4} - 2 β^{2} - 1 = 0

, so

(β^{2} - 1)^{2} = 2

and

β^{2} = 1 + 2

taking the positive root. Hence

α = β^{2} = 2 + 1

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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