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Latus Rectum of Parametric Parabola x=t²+t+1, y=t²+2t+3 | JEE

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Question
The length of the latus rectum of the parabola x=t2+t+1x = t^2 + t + 1, y=t2+2t+3y = t^2 + 2t + 3 is:
A12\dfrac{1}{2}
B12\dfrac{1}{\sqrt{2}}
C122\dfrac{1}{2\sqrt{2}}correct
D18\dfrac{1}{8}
Solution
Step 1: Treat both relations as linear in t2t^2 and tt and cross-multiply.
t22x−y+1=ty−x−2=11⇒t2=2x−y+1,t=−(x−y+2).\frac{t^2}{2x-y+1} = \frac{t}{y-x-2} = \frac{1}{1} \Rightarrow t^2 = 2x-y+1,\quad t = -(x-y+2).
 Step 2: Eliminate tt through t2=(t)2t^2 = (t)^2.
(x−y+2)2=2x−y+1⇒(x−y)2=−2x+3y−3.(x-y+2)^2 = 2x-y+1 \Rightarrow (x-y)^2 = -2x+3y-3.
 Step 3: Rotate to standard form. Add 2λ(x−y)+λ22\lambda(x-y)+\lambda^2 and pick λ\lambda so the right side is perpendicular to x−y+λ=0x-y+\lambda=0:
(−2+2λ)−(3−2λ)=0⇒λ=54,(-2+2\lambda)-(3-2\lambda)=0 \Rightarrow \lambda = \frac{5}{4},
(x−y+54)2=12x+12y−2316=12(x+y−238).\left(x-y+\frac{5}{4}\right)^2 = \frac{1}{2}x + \frac{1}{2}y - \frac{23}{16} = \frac{1}{2}\left(x+y-\frac{23}{8}\right).
 Step 4: Dividing each linear form by 2\sqrt{2} turns it into a perpendicular distance, so P2=122 QP^2 = \dfrac{1}{2\sqrt{2}}\,Q. Comparing with P2=(LR) QP^2 = (\text{LR})\,Q,
LR=122.\text{LR} = \frac{1}{2\sqrt{2}}.
 Correct answer: (3)
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