## anonymous one year ago What is the equation of the quadratic graph with a focus of (4, 0) and a directrix of y = 10?

1. anonymous

@campbell_st

2. campbell_st

well there are lots of ways to do this question. 1. picking a point P(x, y) on the parabola and then finding the distance from the focus to P and equating that to the distance from P to the directrix. 2. using a little observation and a standard form of the parabola $(x - h)^2 = 4a(y - k)$ the vertex is (h, k) and a is the focal length.

3. anonymous

I saw on another of the same problems where someone mentioned that that half-way point between both y's. In this case they are 0 and 10 which would be 5.

4. campbell_st

I always choose option 2... it requires drawing a diagram and you need to know the focal length is the distance between the vertex and focus. It's also the distance between the vertex and directrix. so the distance between the focus and directrix is 2a. so the diagram |dw:1444157997065:dw| ok, so lookin at the diagram the directrix is above the focus, so the parabola is concave down. can you tell me the distance between the focus and directrix?

5. campbell_st

great so you know how to find the focal length, you have a = 5.... so what point is 5 units drectly above the focus...?

6. anonymous

9...?

7. campbell_st

|dw:1444158287455:dw|

8. anonymous

4,5

9. campbell_st

great you have the vertex (4, 5) and you know a = 5 so the equation is now (x - 4)^2 = -4 times 5 (y - 5) remember, the graph has the directrix above the focus, so the parabola is concave down... so thats why its -4. all you need to do is simplify the equation hope it makes sense

10. campbell_st

oops equation should read $(x - 4)^2 = -4 \times 5(y - 5)$

11. campbell_st

so making y the subject you would end up with $y = \frac{-1}{20}(x -4)^2 + 5$