## anonymous one year ago directrix =-2 and focus is at (4,3) find equation and the vertex

1. anonymous

@freckles

2. anonymous

x=-2?

3. anonymous

vertex = -2

4. anonymous

-2? so..hmmm shouldn't it be an ordered pair? 0,-2? -2,0?

5. anonymous

actually i think the directrix is -2

6. anonymous

well then, you found it already then =)

7. anonymous

i need help finding the eq and vertex

8. anonymous

well, the directrix should have an equation x = -2? y = -2?

9. anonymous

x = -2

10. anonymous

|dw:1442276451048:dw| so.. where do you think the vertex would be?

11. anonymous

1,3

12. anonymous

how do i find equation

13. anonymous

well, right.. 1,3 is correct.. the equation is ahemm notice the distance from the vertex to either, the focus or directrix, is 3 units thus p = 3 and it's also a horizontal parabola thus $$\bf \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\qquad thus \\ \quad \\ \begin{array}{llll} (y-{\color{blue}{ 3}})^2=4{\color{purple}{ (3)}}(x-{\color{brown}{ 1}}) \\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 1}},{\color{blue}{ 3}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}$$

14. anonymous

because is a horizontal parabola, the "y" is the squared variable

15. anonymous

would... but need to dash in 1 sec :|

16. anonymous

okay