ParabolamediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Parabola: Let Vertex Parabola Line Through Intersects Circle Points (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let PP be the vertex of the parabola y=x2−6x+12y = x^2-6x+12. If a line through PP intersects the circle x2+y2−2x−4y+3=0x^2+y^2-2x-4y+3=0 at points RR and SS, then the maximum value of (PR+PS)2(PR+PS)^2 is:
A1010
B2020correct
C2525
D55
Solution
Step 1: Find the vertex PP
y=(x−3)2+3  ⟹  P=(3,3)y = (x-3)^2+3 \implies P = (3,3)
 Step 2: Identify the circle's centre CC and radius rr
(x−1)2+(y−2)2=2  ⟹  C=(1,2),r=2(x-1)^2+(y-2)^2 = 2 \implies C=(1,2),\quad r=\sqrt{2}
 Step 3: Verify that PP lies outside the circle
PC=(3−1)2+(3−2)2=5>r=2✓PC = \sqrt{(3-1)^2+(3-2)^2} = \sqrt{5} > r = \sqrt{2} \quad\checkmark
 Step 4: Maximise (PR+PS)2(PR+PS)^2 For a secant through PP at angle θ\theta to PCPC, with foot perpendicular at distance d=PCsin⁡θd = PC\sin\theta from CC:
PR+PS=2 PCcos⁡θPR+PS = 2\,PC\cos\theta
 This is maximised when θ=0\theta=0, i.e., when the chord RSRS passes through CC. At this configuration:
(PR+PS)max⁡=2PC=25  ⟹  (PR+PS)max⁡2=4×5=20(PR+PS)_{\max} = 2PC = 2\sqrt{5} \implies (PR+PS)_{\max}^2 = 4\times5 = 20
 Answer: (2)
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