Question

The length of latus rectum of the parabola whose parametric equations are $x=t^2+t+1$, $y=t^2-t+1$, where $t\in R$, is equal to?

• A. $\sqrt{2}$
• B. $\sqrt{4}$
• C. $\sqrt{8}$
• D. $\sqrt{6}$

Answer 1

The correct option is B $\sqrt{4}$

$x=t^2+t+1$

$y=t^2-t+1$

Adding above equations, we get

$\therefore t=\frac{x-y}{2}$

Substituting this in first equation, we get

$\therefore x={\left(\frac{x-y}{2}\right)}^2+\frac{x-y}{2}+1$

$\therefore \frac{x+y-2}{2}={\left(\frac{x-y}{2}\right)}^2$

$\therefore (x-y{)}^2=2(x+y-2)$

So, length of latus rectum = 2