# Conics: Parabola

• April 5th 2011, 06:20 PM
NeoSonata
Conics: Parabola
Hello! The question is asking me to find the vertex, focus and directrix of the parabola, and sketch its graph.

y= (1/4) (x^2 -2x+5)

I tried putting aside the 1/4 for the time being and tried completing the square

y-5 = (x^2)-2x
y-3 = (x^2)-2x+2

y-3 = (x-1)^2

Plugging back in the 1/4 I got
y-3 = (1/4)(x-1)^2

With that, my vertex is (1, 3)

• April 5th 2011, 06:28 PM
TheChaz
You can't "put aside" the 1/4!!!!!!!!!
Multiply by 4...
4y = x^2 - 2x + 5
4y - 5 = x^2 - 2x
4y - 4 = x^2 - 2x + 1
4(y - 1) = (x - 1)^2
y - 1 = (1/4)(x - 1)^2
y = (1/4)(x - 1)^2 + 1
So (h, k) = (1, 1)
• April 5th 2011, 06:35 PM
TheEmptySet
Quote:

Originally Posted by NeoSonata
Hello! The question is asking me to find the vertex, focus and directrix of the parabola, and sketch its graph.

y= (1/4) (x^2 -2x+5)

I tried putting aside the 1/4 for the time being and tried completing the square

y-5 = (x^2)-2x
y-3 = (x^2)-2x+2

y-3 = (x-1)^2

Plugging back in the 1/4 I got
y-3 = (1/4)(x-1)^2

With that, my vertex is (1, 3)

$y= \left(1/4 \right) (x^2 -2x+5)$ multiply both sides by 4

$4y=x^2-2x+5$ subract 5 from both sides

$4y-5=x^2-2x$ Now complete the square on the x

$4y-5+1=x^2-2x+1=(x-1)^2$

$4(y-1)=(x-1)^2$

Now that this is is standard from we can identify

$4p(y-k)=(x-h)^2$

So $p=1,k=1,h=1$

Vertex $(1,1)$ and the equation of the directrix is $y=-1$

To slow