Question

The parametric representation of a parabola is $x=3+t^2,y=2t-1$ . Its focus is

• A. $(3,-1)$
• B. $(3,0)$
• C. $(4,0)$
• D. $(4,-1)$

Answer 1

Answer: (D)

$(4,-1)$

Given : $x=3+t^2,y=2t-1$

$\to x-3=t^2...\left(1\right)$

$\to \left(\frac{y+1}{2}\right)=t...\left(2\right)$ comparing (1) and (2) we get

$(x-3)=\frac{(y+1{)}^2}{4}\to (y+1{)}^2=4(x-3)$

vertex = $(3,-1)$ , $(x-n)=\frac{1}{4p}(y-k{)}^2$

$\therefore \frac{1}{4p}=\frac{1}{4}\to p=1$

To find focus and directrix first through we should find 'p', since focus is always on the axis of symmetry we need tomove 1 unit from vertex

$\therefore$ focus = $(3+1,-1)\equiv (4,-1)$