Test Index

Parabola Part 2

Section: Mathematics

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Question : 7 of 51

Marks: +1, -0

Let A be the focus of the parabola

y^2=8 x

y=m x+c

intersect the parabola at two distinct points B and C . If the centroid of the triangle ABC is

\left(\; \frac{7}{3},\; \frac{4}{3}\right)

(B C)^2

[28 Jan 2026 Shift 2]

89
80
32
41

Solution:

Concept:

Parametric form on parabola and centroid formula are used to find the parameters of the intersection points.

Explanation:

The parabola

y^2 = 8x

gives

4a = 8

, so

a = 2

.

Focus

A = (2, 0)

.

Take

B = (2t_1^2, 4t_1)

and

C = (2t_2^2, 4t_2)

.

Centroid

G = \left(\frac{7}{3}, \frac{4}{3}\right)

.

From y‑coordinate:

\frac{4t_1 + 4t_2 + 0}{3} = \frac{4}{3}

⇒

t_1 + t_2 = 1

.

From x‑coordinate:

\frac{2t_1^2 + 2t_2^2 + 2}{3} = \frac{7}{3}

⇒

t_1^2 + t_2^2 = \frac{5}{2}

.

Use

(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2

:

1 = \frac{5}{2} + 2t_1t_2

⇒

t_1t_2 = -\frac{3}{4}

.

Then

(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 = 1 - 4\left(-\frac{3}{4}\right) = 4

.

Distance

BC^2 = (2t_2^2 - 2t_1^2)^2 + (4t_2 - 4t_1)^2

= 4(t_2^2 - t_1^2)^2 + 16(t_2 - t_1)^2

= 4(t_2 - t_1)^2(t_2 + t_1)^2 + 16(t_2 - t_1)^2

= 4(t_2 - t_1)^2\left[(t_2 + t_1)^2 + 4\right]

Substitute

(t_2 - t_1)^2 = 4

and

(t_2 + t_1)^2 = 1

:

= 4 \cdot 4 \cdot (1 + 4) = 80

.

Answer:

(BC)^2 = 80

; option B.

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