1. Is my work correct?

Question 1: Find the vertex, focus, and directrix of the parabola:
4x - y^2 - 2y - 33 = 0

My work:
y^2 + 2y = 4x - 33
(y + 1)^2 = 4(x - 8)
Vertex: (8, -1); Focus: (-7, -1); Directrix: x = -9

Question 2: Find the vertex, focus, and directrix of the parabola:
x^2 - 2x + 8y + 9 = 0

My work:
8y = -(x^2 - 2x + 9)
8y = -(x - 1)^2 - 8
8(y + 1) = -(x - 1)^2
Vertex: (1, -1); Focus: (0, -1); Directrix: x = -2

I am not sure if my work is correct, especially in the 2nd question. Can you please tell me if I went wrong?
Thanks!

2. Originally Posted by iluvmathbutitshard
Question 1: Find the vertex, focus, and directrix of the parabola:
4x - y^2 - 2y - 33 = 0

My work:
y^2 + 2y = 4x - 33
(y + 1)^2 = 4(x - 8)
Vertex: (8, -1); Focus: (-7, -1); Directrix: x = -9

Hi iluvmathbutitshard,

I only have time for the first one. Your vertex looks ok, but I'm getting something different for the focus and directrix.

$4x-y^2-2y-33=0$

$4x=y^2+2y+33$

$4x=(y^2+2y+1)+33-1$

$4x=(y+1)^2+32$

$x=\frac{1}{4}(y+1)^2+8$

Vertex $(h, k)=(8, -1)$

Focus $\left(h+\frac{1}{4a}, k\right)$

This would make the focus $\left(8+\frac{1}{4(\frac{1}{4})}, -1\right)=(9, -1)$

And the directrix $x=h-\frac{1}{4a}\Longrightarrow x=8-\frac{1}{4(\frac{1}{4})} \Longrightarrow 8-1 = 7$

3. For the second question, the vertex is correct but the focus and directrix are incorrect. The focus should be two units below the vertex at (1, -3) and the directrix is y = 1.

4. Originally Posted by iluvmathbutitshard

Question 2: Find the vertex, focus, and directrix of the parabola:
x^2 - 2x + 8y + 9 = 0

My work:
8y = -(x^2 - 2x + 9)
8y = -(x - 1)^2 - 8
8(y + 1) = -(x - 1)^2
Vertex: (1, -1); Focus: (0, -1); Directrix: x = -2

I am not sure if my work is correct, especially in the 2nd question. Can you please tell me if I went wrong?
Thanks!
Your vertex looks good to me, but take another look at your Focus and Directrix. Here's my approach.

$x^2-2x+8y+9=0$

$8y=-x^2+2x-9$

$8y=-1(x^2-2x+1)-9+1$

$y=-\frac{1}{8}(x-1)^2-1$

Vertex $(1, -1)$

Focus $\left(h, k+\frac{1}{4a}\right)$

Directrix $y=k-\frac{1}{4a}$

,

,

directrix of parabola x2 8y-2x 13=0

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