## anonymous one year ago Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1

1. campbell_st

have you any method to use for this question...?

2. anonymous

not really no

3. campbell_st

so what methods do you know.. 1. find the focal length and then find the vertex... 2. use the distance formula..

4. anonymous

number 1

5. campbell_st

ok.... so start by plotting the given information |dw:1437680444218:dw| so this tells me the parabola is concave up... does that make sense...

6. anonymous

yes

7. campbell_st

ok... so the line of symmetry is x = -5 and the vertex is a units below the focus... where a = the focal length so |dw:1437680607256:dw|

8. anonymous

so the length between the focus and the directrix is 6

9. campbell_st

the distance between the focus and directrix is 2a the vertex is midway between the vertex and directrix on the line of symmetry

10. campbell_st

that's great so 2a = 6 therefore the focal length is a = 3 is that ok..?

11. anonymous

yes

12. campbell_st

|dw:1437680777548:dw|

13. campbell_st

so where do you think the vertex is located..?

14. anonymous

woud it be at (-5,2)

15. campbell_st

perfect.... now to get the correct equation I use a standard form $(x - h)^2 = 4a(y - k)$ (h, k) is the vertex and a is the focal length so the equation is $(x + 5)^2 = 4 \times 3 (y - 2)$ so this can be simplified and rewritten in the form you need.

16. campbell_st

hope it all makes sense

17. anonymous

Thank you, it really does help

18. campbell_st

if you need it rewritten I'd use $y = \frac{1}{12}(x + 5)^2 + 2$

19. anonymous

That's exactly what I thought it was. Thank you so much