## anonymous one year ago Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.

1. campbell_st

ok... have you plotted the information on the number plane so you know the concavity of the parabola..?

2. anonymous

yeah

3. anonymous

So, you can still use the combined distance equation for this problem. But you just have to make your solution ax^2+bx+c=0.

4. anonymous

ax+by=c.

5. campbell_st

so is the directrix above or below the focus...? and then is the parabola concave up or down..?

6. campbell_st

|dw:1438892282167:dw| so you plot looks like this

7. anonymous

below right?

8. campbell_st

|dw:1438892378795:dw| the distance is labelled 2a and is double the focal length.

9. anonymous

8

10. campbell_st

the dotted line through x = 4 is the line of symmetry.... and the vertex is on this line midway between the focus and directrix

11. campbell_st

great so if the twice the focal length is 8 then the focal length is a = 4 so the vertex is 4 units below the focus on the line x = 4 so where do you think the vertex is..?

12. campbell_st

|dw:1438892601545:dw|

13. anonymous

at y=-11 ??

14. anonymous

or no

15. campbell_st

great... so the standard form I use now is $(x - h)^2 = 4a(y - k)$ (h, k) is the vertex and a is the focal length so the equation is $(x -4)^2 = 4 \times 4(y + 11)$ now you just need to simplify this equation

16. anonymous

f(x)=1/16x^-8x+11 ?

17. campbell_st

so if you want it in vertex form its $\frac{1}{16}(x -4)^2 = y + 11$ then subtract 11 from both sides

18. campbell_st

well not quite $\frac{1}{16}(x^2 - 8x + 16)- 11 = y$

19. campbell_st

which would become $y = \frac{x^2}{16} - \frac{x}{2} - 10$

20. anonymous

ohh ok, thats the same as 1/16x^2-1/2x-10 right?

21. campbell_st

it is the same

22. anonymous

ok thank you so much :)