anonymous one year ago Find the vertex, focus, directrix, and focal width of the parabola. x = 4y^2

1. perl

The general form of your parabola is $$\Large x = \frac {1}{4p} y^2$$Focal width is 4p

2. anonymous

ok what do i do wuth that formula

3. perl

we need to solve 1/(4p) = 4

4. anonymous

oh so p = 1/16?

5. perl

yes

6. perl

the vertex is (h,k) directrix is x = -p focal width is 4p

7. perl

it is centered at the origin

8. anonymous

how do we find h and k

9. anonymous

ok the focal width is .25

10. perl

The general form of a parabola that opens to the right or left $\Large x-h = \frac {1}{4p} (y-k)^2$Vertex is (h,k) directrix is x =h -p focal width is 4p

11. perl

yes that is correct

12. anonymous

is the focal width represented as a single letter?

13. perl

4 times that value of p

14. anonymous

like i know i can plug in x as 4y^2 but thats still not enough to find either h or k

15. perl

h must be zero and k must be zero

16. anonymous

how do you know its at the origin though

17. anonymous

oh wait all the choices have the vertix set to (0,0) but if it isnt how do you know

18. perl

$\Large x-h = \frac {1}{4p} (y-k)^2 \iff x = \frac {1}{4p} y^2$

19. perl

the two equation left sides must match and the right side must match

20. anonymous

ohhhhhh omg ok i get it

21. perl

the two equations left sides must match and right sides

22. anonymous

and thats how you know its at the origin?

23. perl

yes. or by graphing it

24. anonymous

ok then the directrix is 0-.25 which would = -1/4

25. anonymous

26. perl

correct. and the focus is (h+ p, k )

27. anonymous

oh ok this is actually simplier than i thought! thank you so much!