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Chords Bisected by y-axis: 6(α+β) = 3 | JEE Main 2026

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

Question
Suppose the two chords, drawn from the point (1,2)(1,2) on the circle x2+y2+x−3y=0x^2+y^2+x-3y=0, are bisected by the yy-axis. If the other ends of these chords are RR and SS, and the mid point of the line segment RSRS is (α,β)(\alpha,\beta), then 6(α+β)6(\alpha+\beta) is equal to
A11
B33correct
C44
D66
Solution
Step 1: Let a chord from P(1,2)P(1,2) be bisected by the yy-axis at M(0,λ)M(0,\lambda). Then the other end is R or S=(−1, 2λ−2)R\text{ or }S=(-1,\ 2\lambda-2) (using midpoint). Step 2: This end lies on the circle:
(−1)2+(2λ−2)2+(−1)−3(2λ−2)=0 ⇒ 2λ2−7λ+5=0 ⇒ λ=1 or λ=52.(-1)^2+(2\lambda-2)^2+(-1)-3(2\lambda-2)=0\ \Rightarrow\ 2\lambda^2-7\lambda+5=0\ \Rightarrow\ \lambda=1\ \text{or}\ \lambda=\frac52.
 Step 3: For λ=1\lambda=1: end =(−1,0)=(-1,0). For λ=52\lambda=\dfrac52: end =(−1,3)=(-1,3). So R(−1,0)R(-1,0), S(−1,3)S(-1,3) (or vice versa). Step 4: Midpoint of RS=(−1, 32)RS=\left(-1,\ \dfrac32\right), so α=−1, β=32\alpha=-1,\ \beta=\dfrac32, and
6(α+β)=6(−1+32)=6⋅12=3.6(\alpha+\beta)=6\left(-1+\frac32\right)=6\cdot\frac12=3.
 Correct answer: (2)
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