For the parabola represented in the parameteric form by $x=t^2+t+1$ and $y=t^2-t+1$, the length of latus rectum is

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Aicharya · Accepted Answer

Given, $\begin{aligned} & x=t^2+t+1 \ & y=t^2-t+1 \Rightarrow x+y=2\left(t^2+1ight)\end{aligned}$ and $\quad(x-y)=2 t$ ...(i) $\Rightarrow \quad t=\frac{x-y}{2}$ ...(ii) From Eqs. (i) and (ii), $x+y=2\left[\frac{(x-y)^2}{4}+1ight]$ $\begin{array}{ll}\Rightarrow & 2(x+y)=(x-y)^2+4 \ \Rightarrow & (x-y)^2=2(x+y-2)\end{array}$ $Y^2=2 X$ where, $\quad Y=x-y$ and $\quad X=x+y-2$ $	herefore$ Length of latus rectum $=2$

For the parabola represented in the parameteric form by $x=t^2+t+1$ and $y=t^2-t+1$, the length of latus…

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