JEE Main 2025 Jan 24 Shift 2, Mathematics Q20: Introduction to Parabola

The axis of the parabola is perpendicular to the directrix $x+2y=0$ (slope $-1/2$). So the axis has slope $2$ and passes through $V(3/2,3)$:

$y-3=2\left(x-\frac{3}{2}\right)\Rightarrow y=2x(\text{i.e. }y-2x=0)$

The foot of the directrix on the axis is the intersection of $y-2x=0$ with $x+2y=0$, which gives $x=0,y=0$, i.e. the origin.

The vertex bisects the segment from the foot of the directrix to the focus, so the focus is the reflection of $(0,0)$ in $V$:

$F=(2\cdot 3/2,2\cdot 3)=(3,6)$

Apply the focus-directrix definition $PS=PM$ (distance to focus = distance to directrix):

$(x-3{)}^2+(y-6{)}^2={\left(\frac{x+2y}{\sqrt{5}}\right)}^2$

Multiply through by $5$:

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

$5(x^2-6x+9+y^2-12y+36)=x^2+4xy+4y^2$

$5x^2+5y^2-30x-60y+225=x^2+4xy+4y^2$

$4x^2+y^2-4xy-30x-60y+225=0$

Matching: $\alpha =4,\beta =1,\gamma =4$. So $\alpha +\beta +\gamma =9$.

Correct option: (4)