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

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

Question

The length of the latus rectum of the parabola

x = t^2 + t + 1

y = t^2 + 2t + 3

is:

\dfrac{1}{2}

\dfrac{1}{\sqrt{2}}

\dfrac{1}{2\sqrt{2}}

correct

\dfrac{1}{8}

Solution

Step 1: Treat both relations as linear in

t^2

and

t

and cross-multiply.

\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

t

through

t^2 = (t)^2

(x-y+2)^2 = 2x-y+1 \Rightarrow (x-y)^2 = -2x+3y-3.

Step 3: Rotate to standard form. Add

2\lambda(x-y)+\lambda^2

and pick

\lambda

so the right side is perpendicular to

x-y+\lambda=0

(-2+2\lambda)-(3-2\lambda)=0 \Rightarrow \lambda = \frac{5}{4},

\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

\sqrt{2}

turns it into a perpendicular distance, so

P^2 = \dfrac{1}{2\sqrt{2}}\,Q

. Comparing with

P^2 = (\text{LR})\,Q

\text{LR} = \frac{1}{2\sqrt{2}}.

Correct answer: (3)

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