## tw101 one year ago Write the equation of the parabola with a focus (1,2) and directrix y = 1.

1. tw101

Im not sure how to start.

2. tw101

1?

3. tw101

okay the formula you put up above?

4. tw101

oh okay, onne second and then ill tell you what I got:)

5. jdoe0001

one may note that 1,2 is the focus point, not the vertex point

6. tw101

oh wait what does that mean then?

7. jdoe0001

well... she maybe crazy, but still beautiful, so tis ok =)

8. tw101

haha.

9. tw101

im seriously confused though...

10. jdoe0001

|dw:1439942544203:dw|

11. tw101

Im not sure... Im sorry, this is new to me.

12. jdoe0001

hmmm well.. have you covered parabolas yet?

13. tw101

Yes, briefly.

14. jdoe0001

well.. it turns out that the focus point is "p" distance from the vertex and the directrix is "p" distance also from the vertex so.. the vertex is really half-way between both, since the focus and directrix are the same distance from the vertex keeping in mind that, the parabola opens up towards the focus point

15. tw101

Okay, well crazyandbeautiful helped me out and said that (1,3/2) was the vertex. haha.

16. jdoe0001

hmmm how did she get 1, 3/2 though?

17. tw101

haha no you dont need to its okay! @Crazyandbeautiful

18. tw101

okay so its always halfway between the focus point and directrix?

19. jdoe0001

yeap, they're always "p" distance from the vertex

20. tw101

Okay, i get that now:)

21. tw101

Where do I go from there?

22. jdoe0001

so... get the vertex, get "p" and plug and chug $$\large{ \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}}$$