Question

The parametric equations of a parabola are $x=t^2+1,y=2t+1$.The cartesian equation of its directrix is

• A. x=0
• B. x+1=0
• C. y=0
• D. none of these

Answer 1

The correct option is A

x=0

Given:

$x=t^2+1$ ...(1)

$y=2t+1$ ...(2)

From (!) and (2) :

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

On simplifying : $(y-1{)}^2=4(x-1)$

Let Y=y-1 and X=x-1

$\therefore Y^2=4X$

Comparing it with $y^2=4aX$

a=1

Therefore,the equation of the directrix is X=-a,i.e. $x-1=-1\Rightarrow x=0$