JEE Main 2025 Jan 29 Shift 2, Mathematics Q25: Circle: Chord, Tangent And Normal

For $y^2=12x$, $4a=12\Rightarrow a=3$ and focus $S=(3,0)$. Parameterise $P=(3t^2,6t)$, $Q=(3/t^2,-6/t)$ since $t_1{t}_2=-1$ on a focal chord.

Focal distances: $SP=3+3t^2=3(1+t^2)$, $SQ=3(1+1/t^2)$.

$SP\cdot SQ=9\cdot \frac{(1+t^2{)}^2}{t^2}=\frac{147}{4}\Rightarrow \frac{(1+t^2{)}^2}{t^2}=\frac{49}{12}$.

Letting $u=t^2$, $12(1+u{)}^2=49u\Rightarrow 12u^2-25u+12=0\Rightarrow u=\frac{4}{3}$ or $\frac{3}{4}$.

Choose $t^2=\frac{3}{4}$ (take $t=-\frac{\sqrt{3}}{2}$):

$P=\left(\frac{9}{4},-3\sqrt{3}\right)$, $Q=(4,4\sqrt{3})$.

Circle on $PQ$ as diameter:

$(x-\frac{9}{4})(x-4)+(y+3\sqrt{3})(y-4\sqrt{3})=0$

Expanding: $x^2-\frac{25}{4}x+9+y^2-\sqrt{3}y-36=0$

$\Rightarrow x^2+y^2-\frac{25}{4}x-\sqrt{3}y-27=0$.

Multiplying by $64$: $64x^2+64y^2-400x-64\sqrt{3}y-1728=0$, i.e. $64x^2+64y^2-400x-64\sqrt{3}y=1728$.

Comparing, $\alpha =400$, $\beta =1728$, so $\beta -\alpha =1328$.