Parabola – Focal Chord & Diametric Circle – JEE Main 29 Jan 2025 Shift 2

Solution:

For the parabola $y^2=12x$, comparing with $y^2=4ax$, we get $4a=12 \Rightarrow a=3$. The focus $S$ is $(3,0)$.

Let $P(at^2, 2at)$ and $Q(a/t^2, -2a/t)$ be the endpoints of the focal chord $PQ$.
The focal distances are given by $SP = a(1+t^2)$ and $SQ = a(1+\frac{1}{t^2})$.

Given $SP \cdot SQ = \frac{147}{4}$:
$$ a^2 (1+t^2) \left(1+\frac{1}{t^2}\right) = \frac{147}{4} $$
$$ 9 \frac{(1+t^2)^2}{t^2} = \frac{147}{4} \Rightarrow \frac{(1+t^2)^2}{t^2} = \frac{49}{12} $$
$$ 12(1+t^2)^2 = 49t^2 $$
$$ 12(t^4 + 2t^2 + 1) = 49t^2 \Rightarrow 12t^4 – 25t^2 + 12 = 0 $$
$$ (4t^2 – 3)(3t^2 – 4) = 0 \Rightarrow t^2 = \frac{3}{4} \text{ or } \frac{4}{3} $$

We take $t^2 = \frac{3}{4}$, so $t = -\frac{\sqrt{3}}{2}$ (choosing the negative root consistent with the problem context).
Coordinates of $P$:
$$ P\left(3\left(\frac{3}{4}\right), 6\left(-\frac{\sqrt{3}}{2}\right)\right) = \left(\frac{9}{4}, -3\sqrt{3}\right) $$
Coordinates of $Q$:
$$ Q\left(3\left(\frac{4}{3}\right), -6\left(-\frac{2}{\sqrt{3}}\right)\right) = \left(4, 4\sqrt{3}\right) $$

Equation of the circle with diameter $PQ$:
$$ (x – \frac{9}{4})(x – 4) + (y + 3\sqrt{3})(y – 4\sqrt{3}) = 0 $$
$$ x^2 – 4x – \frac{9}{4}x + 9 + y^2 – \sqrt{3}y – 36 = 0 $$
$$ x^2 + y^2 – \frac{25}{4}x – \sqrt{3}y – 27 = 0 $$

Multiply by 64 to match the given form:
$$ 64x^2 + 64y^2 – 400x – 64\sqrt{3}y – 1728 = 0 $$
$$ 64x^2 + 64y^2 – 400x – 64\sqrt{3}y = 1728 $$

Comparing with $64x^2 + 64y^2 – \alpha x – 64\sqrt{3}y = \beta$:
$$ \alpha = 400, \quad \beta = 1728 $$
$$ \beta – \alpha = 1728 – 400 = 1328 $$

Ans. 1328