JEE Main 2026 — Parabola Question with Solution

The equation of the parabola is $y=x^2-6x+12$, which can be rewritten as $y-3=(x-3{)}^2$.

The vertex of the parabola is $P(3,3)$.

The equation of the circle is $x^2+y^2-2x-4y+3=0$, which can be rewritten as $(x-1{)}^2+(y-2{)}^2=2$.

The center of the circle is $C(1,2)$ and its radius is $r=\sqrt{2}$.

The distance between the point $P$ and the center $C$ is:

$PC=\sqrt{(3-1{)}^2+(3-2{)}^2}=\sqrt{4+1}=\sqrt{5}$

Since $PC=\sqrt{5}>r=\sqrt{2}$, the point $P$ lies outside the circle.

Let the line through $P$ intersect the circle at $R$ and $S$, and let $M$ be the midpoint of the chord $RS$. The perpendicular distance from the center $C$ to the chord is $CM$.

Using the right-angled triangle $△PCM$, the distance from $P$ to the midpoint $M$ is $PM=\sqrt{P{C}^2-C{M}^2}$.

The lengths of the segments are $PR=PM-RM$ and $PS=PM+RM$.

Therefore, the sum of the distances is:

$PR+PS=(PM-RM)+(PM+RM)=2PM=2\sqrt{P{C}^2-C{M}^2}$

To maximize $(PR+PS{)}^2$, we need to maximize $P{C}^2-C{M}^2$. This occurs when $CM$ is minimized, which means $CM=0$. This happens when the line passes through the center $C$ of the circle.

Substituting $CM=0$, we get the maximum sum:

$PR+PS=2PC=2\sqrt{5}$

Thus, the maximum value of $(PR+PS{)}^2$ is:

$(2\sqrt{5}{)}^2=4\times 5=20$

Answer: $20$