If the equation of the parabola with vertex $\mathrm{V}\left(\frac{3}{2}, 3\right)$ and the directrix $\mathrm{x}+2 \mathrm{y}=0$ is $\alpha x^{2}+\beta y^{2}-\gamma x y-30 x-60 y+225=0$, then $\alpha+\beta+\gamma$ is equal to:

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Parabola

If the equation of the parabola with vertex

V (\frac{3}{2}, 3)

and the directrix

x + 2 y = 0

α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0

, then

α + β + γ

is equal to:

Answer: (D)

The axis of the parabola is perpendicular to the directrix

x + 2 y = 0

, whose normal direction is

(1, 2)

; so the axis through the vertex

V (\frac{3}{2}, 3)

has slope 2:

y - 3 = 2 (x - \frac{3}{2}) \Rightarrow y = 2 x .

The foot of the perpendicular from

V

to the directrix lies on both this axis and the directrix. Solving

y = 2 x

with

x + 2 y = 0

x + 4 x = 0 \Rightarrow x = 0, y = 0

— the foot is the origin.

Since the vertex is the midpoint of the focus

F

and this foot,

F = 2 V - (0, 0) = (3, 6)

.

For any point

(x, y)

on the parabola, distance to focus equals distance to directrix:

(x - 3)^{2} + (y - 6)^{2} = (\frac{x + 2 y}{5})^{2} .

Multiplying by 5 and expanding:

5 (x - 3)^{2} + 5 (y - 6)^{2} = (x + 2 y)^{2}

5 x^{2} - 30 x + 45 + 5 y^{2} - 60 y + 180 = x^{2} + 4 x y + 4 y^{2}

4 x^{2} + y^{2} - 4 x y - 30 x - 60 y + 225 = 0.

Comparing with

α x^{2} + β y^{2} - γ x y - 30 x - 60 y + 225 = 0

α = 4, β = 1, γ = 4

, so

α + β + γ = 4 + 1 + 4 = 9.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option