JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Parabola

Let R be the focus of the parabola

y^{2} = 20 x

and the line

y = mx + c

intersect the parabola at two points P and Q. Let the point

G (10, 10)

be the centroid of the triangle PQR. If

c - m = 6

, then

(PQ)^{2}

Answer: (A)

Solution

For

y^{2} = 20 x

4 a = 20

a = 5

and the focus is

R = (5, 0)

. Write

P = (5 t_{1}^{2}, 10 t_{1})

and

Q = (5 t_{2}^{2}, 10 t_{2})

.

The centroid of

△ P QR

(\frac{5 t _{1}^{2} + 5 t _{2}^{2} + 5}{3}, \frac{10 t _{1} + 10 t _{2}}{3}) = (10, 10)

. From the ordinate,

t_{1} + t_{2} = 3

; from the abscissa,

t_{1}^{2} + t_{2}^{2} = 5

. Then

2 t_{1} t_{2} = (t_{1} + t_{2})^{2} - (t_{1}^{2} + t_{2}^{2}) = 9 - 5 = 4

, so

t_{1} t_{2} = 2

.

Thus

t_{1}, t_{2}

are the roots of

t^{2} - 3 t + 2 = 0

, namely

t_{1} = 1, t_{2} = 2

, giving

P = (5, 10)

and

Q = (20, 20)

. These yield slope

m = \frac{2}{t _{1} + t _{2}} = \frac{2}{3}

and

c = m + 6 = \frac{20}{3}

, consistent with

c - m = 6

.

Therefore

(P Q)^{2} = (20 - 5)^{2} + (20 - 10)^{2} = 225 + 100 = 325.

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option