## anonymous one year ago What is the equation of the quadratic graph with a focus of (2, 0) and a directrix of y = −12?

1. jdoe0001

|dw:1437691840404:dw| any ideas where those two fellas are in the graph? can you graph them there?

2. anonymous

|dw:1437691171977:dw|

3. anonymous

right?

4. jdoe0001

yeap we also know that the vertex is equidistant from both meaning, the focus point is away from the vertex, the same amount the directrix is away from the vertex let's call that distance "p" so, that means the vertex is "half-way" between the focus and directrix the parabola opens towards the focus, the focus is above the directrix, in this case the parabola opens upwards then |dw:1437692110933:dw|

5. jdoe0001

the distance from the focus to the vertex is 7, the distance from the directrix to the vertex is 7 the parabola is opening upwards, so the distance "p" is positive the squared variable is the "x" then now that we know where the vertex coordinates are at, and the "p" distance then $$\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$ plug in your values, and solve for "y"

6. anonymous

wouldn't the focus be on the x-axis, not the y-axis? since it is (2,0) not (0,2)

7. jdoe0001

hmmmm hold the mayo.. .maybe .... I did mess up there, lemme recheck that

8. jdoe0001

|dw:1437692548026:dw|

9. jdoe0001

anyhow, the vertex is half-way, so that makes it a 0,-6 and yes, that makes the "p" distance to 6 only $$\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$

10. jdoe0001

hmmm wait a sec.. dohh. the x coordinate can't be 0 anyhow hehe one sec

11. jdoe0001

|dw:1437692768286:dw| $$\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 2}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$

12. anonymous

ok, we have the vertex, now what do we do to get the equation for it

13. jdoe0001

just plug in the values in the parabola equation focus form

14. anonymous

so it ends up as (x-2)^2 = 4*6(y-(-6))

15. jdoe0001

yeap or $$\bf (x-{\color{brown}{ 2}})^2=4{\color{purple}{ (6)}}(y-{\color{blue}{ (-6)}})\implies (x-2)^2=24(y+6) \\ \quad \\ (x-2)^2=24y+144\implies \cfrac{(x-2)^2-144}{24}=y$$

16. anonymous

Is there another way to write that out, because none of the above are answer choices

17. jdoe0001

I guess one could distribute the denominator, and end up with $$\cfrac{(x-2)^2-144}{24}=y\implies \cfrac{(x-2)^2}{24}-\cfrac{\cancel{144}}{\cancel{24}}=y\implies \cfrac{(x-2)^2}{24}-6=y$$

18. jdoe0001

you could also, expand the squared binomial, and simplify it with the 144, and then distribute the denominator

19. anonymous

my anwer choices are x^2-4-35; -x^2+4+35; -1/24x^2+1/6x-35/6; and 1/24x^2-1/6-35/6

20. jdoe0001

ok one sec

21. jdoe0001

$$\bf \cfrac{(x-2)^2-144}{24}=y\implies \cfrac{[x^2-4x+4]-144}{24}=y \\ \quad \\ \cfrac{x^2-4x+4-144}{24}=y\implies \cfrac{x^2-4x-140}{24}=y \\ \quad \\ \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y\implies ?$$

22. anonymous

what?

23. jdoe0001

any... thing.. that confuses there?

24. anonymous

what happens to the second and third fractions

25. jdoe0001

yeah... whatever happens to those? so... simplify them, what do you get for the simplification? or 4/24 and 140/24?

26. anonymous

oh, ok. The would simplify to 2/12 and 35/8

27. anonymous

sorry 1/6

28. jdoe0001

well... 2/12 = 1/6 and.... 140/24 simplifiefs to 35/6 thus $$\bf \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y \\ \quad \\ \cfrac{1}{24}x^2-\cfrac{1}{6}x-\cfrac{35}{6}=y$$

29. anonymous

oh, ok. Thanks for the help

30. jdoe0001

yw