Oblique parabola – JEE Main 24 January 2025 Shift 2

Solution:

Equation of axis passing through $V(\frac{3}{2}, 3)$ and perpendicular to directrix $x+2y=0$:
$$ y-3 = 2\left(x-\frac{3}{2}\right) $$
$$ y-3 = 2x-3 \Rightarrow y-2x=0 $$

Foot of perpendicular from focus to directrix (Intersection of $y-2x=0$ and $x+2y=0$):
$$ \Rightarrow (0,0) $$

Let Focus be $S(h, k)$. Since vertex $V$ is the midpoint of $S$ and the foot of the directrix:
$$ \frac{h+0}{2} = \frac{3}{2} \Rightarrow h=3 $$
$$ \frac{k+0}{2} = 3 \Rightarrow k=6 $$
Focus $S = (3, 6)$.

Equation of parabola ($PS^2 = PM^2$):
$$ (x-3)^2 + (y-6)^2 = \left( \frac{x+2y}{\sqrt{5}} \right)^2 $$
$$ 5(x^2 – 6x + 9 + y^2 – 12y + 36) = x^2 + 4y^2 + 4xy $$
$$ 5x^2 + 5y^2 – 30x – 60y + 225 = x^2 + 4y^2 + 4xy $$
$$ 4x^2 + y^2 – 4xy – 30x – 60y + 225 = 0 $$

Comparing with $\alpha x^2 + \beta y^2 – \gamma xy – 30x – 60y + 225 = 0$:
$$ \alpha = 4, \quad \beta = 1, \quad \gamma = 4 $$
$$ \alpha + \beta + \gamma = 4 + 1 + 4 = 9 $$

Ans. (4)