Question

Find the vertex, axis, focus, directrix,lastusrectum of the parabola. $x^2-2x+4y+9=0$

Answer 1

$x^2-2x+4y+9=0$

$\therefore (x^2-2x+1)+4y+8=0$

$\therefore {(x-1)}^2+4(y+2)=0$

$\therefore {(x-1)}^2=-4(y+2)$

$\therefore {(x-1)}^2=4(-y-2)$

$\therefore$ Vertex :- ${(x-1)}^2=0$

$(-y-2)=0$

$\therefore$ Vertex:- \left(1,-2\right)

Axis ${(x-1)}^2=0$

$\therefore x=1$

Direction :- $-y-2=-1$

$\therefore y=-1$

Latus rectum $4$