JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Parabola

Let

A, B

and C be three points on the parabola

y^{2} = 6 x

and let the line segment

A B

meet the line

L

through C parallel to the x-axis at the point D. Let M and N respectively be the feet of the perpendiculars from A and B on L.

Then

(\frac{AM \cdot BN}{CD})^{2}

is equal to ____.

Answer: 36

Solution

Parametrise

y^{2} = 6 x

(so

4 a = 6, a = \frac{3}{2}

) as

A = (a t_{1}^{2}, 2 a t_{1})

B = (a t_{2}^{2}, 2 a t_{2})

C = (a t_{3}^{2}, 2 a t_{3})

;

L

is the line

y = 2 a t_{3}

M

, the foot of perpendicular from

A

L

, is

(a t_{1}^{2}, 2 a t_{3})

, so

A M = 2 a ∣ t_{1} - t_{3} ∣

. Similarly

B N = 2 a ∣ t_{2} - t_{3} ∣

.

Chord

A B

has equation

y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}

. Setting

y = 2 a t_{3}

to find

D

(intersection with

L

x_{D} = a t_{3} (t_{1} + t_{2}) - a t_{1} t_{2}

C D = ∣ x_{D} - a t_{3}^{2} ∣ = a ∣ t_{3} t_{1} + t_{3} t_{2} - t_{1} t_{2} - t_{3}^{2} ∣ = a ∣ t_{3} (t_{1} - t_{3}) - t_{2} (t_{1} - t_{3}) ∣ = a ∣ t_{1} - t_{3} ∣∣ t_{2} - t_{3} ∣

Therefore

\frac{A M \cdot B N}{C D} = \frac{4 a ^{2} ∣ t _{1} - t _{3} ∣∣ t _{2} - t _{3} ∣}{a ∣ t _{1} - t _{3} ∣∣ t _{2} - t _{3} ∣} = 4 a

(\frac{A M \cdot B N}{C D})^{2} = 16 a^{2} = 16 (\frac{3}{2})^{2} = 36

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer