## chrisdbest one year ago Help!!

1. chrisdbest

Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

2. chrisdbest

@pooja195

3. chrisdbest

@Nnesha @Preetha

4. jdoe0001

so.. the vertex is a the origin the focus point is a 0. -7 care to draw those two in a cartesian plane?|dw:1437175007773:dw|

5. chrisdbest

Vertex is at origin, and focus point is at 0, -7 like you said

6. jdoe0001

ok... so what are the coordinates for the origin?

7. chrisdbest

Isn't it 0,0?

8. jdoe0001

yeap so |dw:1437175183982:dw| so.... if the parabola makes a U-turn at the 0,0 and the focus is below the vertex that means the parabola is |dw:1437175290874:dw|

9. jdoe0001

so is a parabola that opens vertically meaning the "x" term is the one that's squared so $$\bf \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\ \quad \\\\ \quad \\ \textit{so we'd need to use this form }(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})$$

10. jdoe0001

now... what is the distance "p", that is, from the vertex to the focus point?

11. chrisdbest

hmmm

12. jdoe0001

|dw:1437175751363:dw|

13. chrisdbest

Wouldn't that be 7?

14. jdoe0001

yeap so p is 7 the parabola is going downards, that means "p" is negative so "p" is really negative 7 or -7 so now you know what "p" is, and where the vertex is at, that is what "h" and "k" are so just plug them in at $$\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ 7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$ and simplify and solve for "y" :)

15. jdoe0001

well... negative 7 rather $$\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$

16. chrisdbest

Ohk, yeah, I meant down 7

17. chrisdbest

So what do I do next?

18. jdoe0001

you shall... plug in those values simplify and solve for "y" :)

19. chrisdbest

what values?

20. chrisdbest

So y^2 = -28x?

21. chrisdbest

@jdoe0001

22. jdoe0001

$$\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$ notice.... you have all the values you need, just plug them in and solve for "y"

23. chrisdbest

so what's h?

24. jdoe0001

look at the colors, h,k are just the vertex coordinates

25. chrisdbest

ok

26. chrisdbest

Then what does x equal?

27. chrisdbest

Ohhh. wait I think I see

28. chrisdbest

My answer choices are... A. $y = -\frac{ 1 }{ 7 }x^2$ B. $y^2 = -7x$ C. $y=-\frac{ 1 }{ 28 }x^2$ D. $y^2 = -28x$

29. chrisdbest

So I would choose D?

30. chrisdbest

OMG!!! It's not D!!!

31. dumbcow

Answer is C note the parabola opens down, so its of the form y = x^2 the "y" is squared only if parabola is sideways or opens left/right