JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Parabola

If the three normals drawn to the parabola,

y^{2} = 2 x

pass through the point

(a, 0) a \neq = 0

, then '

a

' must be greater than :

(A) $\frac{1}{2}$
(B)1✓ Correct
(C) $- \frac{1}{2}$
(D)-1

Answer: (B)

Solution

Writing

y^{2} = 2 x

y^{2} = 4 A x

gives

A = \frac{1}{2}

, and the normal at the point

(A t^{2}, 2 A t)

y + t x = 2 A t + A t^{3} = t + \frac{t ^{3}}{2}

Putting

(a, 0)

in it gives

t a = t + \frac{t ^{3}}{2}

, that is

t (\frac{t ^{2}}{2} + 1 - a) = 0

Besides

t = 0

, two further real parameters arise only when

t^{2} = 2 (a - 1) > 0

, so three distinct normals pass through

(a, 0)

only when

a > 1

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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