Q. Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 4/3), then (BC)² is equal to :

Correct Answer: 80

Explanation

The given parabola is

y² = 8x

Comparing with y² = 4ax, we get

a = 2

Hence, the focus of the parabola is

A(2, 0)

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Let the line y = mx + c intersect the parabola at points B(x₁, y₁) and C(x₂, y₂).

Substitute y = mx + c in y² = 8x :

(mx + c)² = 8x

m²x² + 2mcx + c² − 8x = 0

So, the x-coordinates of B and C are the roots of this quadratic.

Using properties of roots,

x₁ + x₂ = (8 − 2mc)/m²
x₁x₂ = c²/m²

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The centroid of triangle ABC is

\left( \dfrac{2 + x₁ + x₂}{3}, \dfrac{0 + y₁ + y₂}{3} \right)

Given centroid is

\left( \dfrac{7}{3}, \dfrac{4}{3} \right)

Equating x-coordinates,

2 + x₁ + x₂ = 7 ⟹ x₁ + x₂ = 5

Equating y-coordinates,

y₁ + y₂ = 4

But

y₁ + y₂ = m(x₁ + x₂) + 2c

4 = 5m + 2c

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Now,

(BC)² = (x₁ − x₂)² + (y₁ − y₂)²

= (x₁ − x₂)²(1 + m²)

Using

(x₁ − x₂)² = (x₁ + x₂)² − 4x₁x₂

= 25 − 4(c²/m²)

Using the earlier relations, simplifying gives

(BC)² = 80

Hence, the required value is

80

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