JEE Main 2025 Apr 4 Shift 2, Mathematics Q7: Introduction to Parabola

Along the line $y=x$, points at distance $\sqrt{2}$ and $2\sqrt{2}$ from the origin (in the first quadrant) are the vertex $V=(1,1)$ and the focus $S=(2,2)$.

The directrix is perpendicular to the axis (so it has slope $-1$) and the vertex is the midpoint between the focus and the foot of perpendicular on the directrix. The foot of perpendicular from $V$ on the directrix is $(0,0)$, giving the directrix as $x+y=0$.

For any point $P(x,y)$ on the parabola, $PS=$ distance from $P$ to the directrix:

$\sqrt{(x-2{)}^2+(y-2{)}^2}=\frac{|x+y|}{\sqrt{2}}$.

Substitute $(1,k)$:

$(1-2{)}^2+(k-2{)}^2=\frac{(1+k{)}^2}{2}$

$2[1+(k-2{)}^2]=(1+k{)}^2$

$2+2k^2-8k+8=k^2+2k+1$

$k^2-10k+9=0\Rightarrow (k-1)(k-9)=0$.

So $k=1$ or $k=9$. The possible value listed is $\mathbf{k}=9$.