Question

How do you find the vertex, focus and directrix of `y^2-2y-2x+5=0`?

Answer 1

The vertex is `=(2,1)`
The focus is `=(2.5,1)`
The directrix is `x=1.5`

Explanation:

Let's rewrite the equation of the parabola
`y^2-2y+1-2x+4=0`
`(y-1)^2=2(x-2)`
This is the equation in the standard form
`(y-b)^2=2p(x-a)`
So the vertex is `(a,b)` `=>` `(2,1)`
The focus is `(a+p/2,b)` `=>` `(2.5,1)`
And the directrix is `x=a-p/2` `=>` `x=2-0.5=1.5`

graph{(y-1)^2=2(x-2) [-5.75, 6.734, -1.29, 4.95]}

Answer 2

The vertex is V( 2, 1), the focus is S(2.5, 1) and the directrix

DX is x=1.5.

Explanation:

The focus S is on the axis VS of the parabola, at a distance ( size of

the parabola ) a, from the vertex V. The directrix DX is perpendicular

to the axis VS, at a distance a, on the opposite side

In respect of the parabola `(y -beta )^2=4a ( x-alpha )`,

the size parameter is a,

the vertex V is `( alpha, beta )`,

the axis VS perpendicular to the directrix DX is `y = beta`,

the tangent VT at the vertex is `x = alpha`,

the directrix DX is `x = alpha -a` and

the focus S is `( alpha+a, beta)`.

Here, the equation is `( y-1 )^2=2( x-2 )`, and so,

`a =1/2, alpha=2 and beta = 1`.

Thus the vertex is #V( 2, 1), the focus is S(2.5, 1) and the directrix

DX is `x=1.5`.