A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = 5.
anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = 5.

This Question is Closed

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1have you sketched the focus and directrix to determine which way the parabola opens...?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1there are 2 ways to do this question... 1. use the distance formula... knowing the distance from the focus to a point on the parabola is equal to the distance from the point to the directirx. 2. find the focal length and use a standard form of the parabola.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1great so the standard from I use for this parabola is \[(y  k)^2 = 4a(x  h)\] where (h, k) is the vertex and a is the focal length. any ideas on the focal length..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea what that is.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just remembered what that was

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok... so it seems you have only been taught about selecting a point on the parabola.... dw:1436392195951:dw do by definition, the distance from the focus to P is equal to the distance from P to Q I'd expect that's the way you have been taught to find the locus of the parabola.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436392369178:dw so which method do you want to use..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the line of symmetry for the parabola is y = 0 the vertex is halfway between the directrix and focus.. so any ideas of the location of the vertex and the focal length..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont care. Whichever you are better with or think is easier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Vertex is (0,0) and focal length is 5

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok... so how far between the directrix and focus?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1great so 2a = 10 so a = 5 is that ok...?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the vertex in this question is to the left of the focus... and it is the focal length to the left on the line of symmetry any ideas in where the vertex is..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1great so you know h = 0, k =0 and a = 5 plug them into the standard form \[(y  k)^2 = 4a(x  h)\] (h, k) is the vertex and a is the focal length
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.