Find the equation in standard form of the parabola?

Question

Focus (3,4) .. and a directrix y = 1


show process please

Answer 1

Using the vertex form of the parabola, y - k = a(x - h)²,
the focus is (h, k + 1/(4a)) and the directrix is y = k - 1/(4a).

Therefore, (h, k + 1/(4a)) = (3,4) in this problem. h = 3 and we can solve for k using k + 1/(4a) = 4 and k - 1/(4a) = 1 (simultaneous equations). Add these two equations together and the terms with a cancel out. You are left with 2k = 5 so then k must be 2.5. Plug this into either equation to find a and you find that (2.5 + 1/(4a) = 4 => 1/(4a) = 1.5 =>4a = 1/1.5 => 1/6) a = 1/6.

Your equation then is y = (1/6)(x - 3)² +2.5.

Answer 2

The standard form of the parabola is y=a(x-h)^2+k (if it opens upward or downwards, which is the case in this problem).
The equation for vertex is (h,k), focus is (h, k+1/4a), and directrix is y=k-1/4a.

Looking at the equations for the focus and directrix, you automatically know that h is 3. Then, k+1/4a must equal 4 and k-1/4a must equal to 1 (your focus and directrix).

If you add these two equations together, it comes out to 2k=5. Thus, k=2.5.

You can now plug k into either the focus or directrix equation.

If you plug it into the focus equation, it's (3, 2.5+1/4a)=(3, 4). 2.5+1/4a=4
1/4a=1.5 so a=1/6.
If you plug it into the directrix equation, it's y=2.5-1/4a=1.
-1/4a=-1.5 so a=1/6.

All that's left to do is plug in your numbers. So, your final equation would be y=1/6(x-3)^2+2.5.

Answer 3

Since the directrix is a horizontal line the parabola opens vertically. The vertex is midway between the focus and the directrix. The vertex (h,k) is:

(h,k) = [3, (4 + 1)/2] = (3, 5/2)

So the equation of the parabola is:

4p(y - k) = (x - h)²
4p(y - 5/2) = (x - 3)²

The directed distance from the vertex to the focus is p.

p = 4 - 5/2 = 3/2
4p = 6

4p(y - 5/2) = (x - 3)²
6(y - 5/2) = (x - 3)²

Answer 4

vertex is halfway between focus and directrix, so v(3, 2.5). also, with horizontal directrix with focus above it, parabola opens up, so

4a(y-y0) = (x-x0)², with x0 = 3, y0 = 2.5, and a is distance between vertex and focus, 1.5, so

6(y - 4) = (x - 3)² or
y = (1/6)(x - 3)² + 4