anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Center is (0,0)
anonymous
  • anonymous
@campbell_st
campbell_st
  • campbell_st
have you sketched the focus and directrix to determine which way the parabola opens...?

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anonymous
  • anonymous
It opens right
campbell_st
  • campbell_st
there 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
  • campbell_st
great 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
  • anonymous
I have no idea what that is.
anonymous
  • anonymous
wait. it's 5
anonymous
  • anonymous
I just remembered what that was
campbell_st
  • campbell_st
ok... 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
  • campbell_st
|dw:1436392369178:dw| so which method do you want to use..?
campbell_st
  • campbell_st
the 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
  • anonymous
I dont care. Whichever you are better with or think is easier.
anonymous
  • anonymous
Vertex is (0,0) and focal length is 5
campbell_st
  • campbell_st
ok... so how far between the directrix and focus?
anonymous
  • anonymous
10
campbell_st
  • campbell_st
great so 2a = 10 so a = 5 is that ok...?
anonymous
  • anonymous
yup
campbell_st
  • campbell_st
the 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..?
anonymous
  • anonymous
0,0?
campbell_st
  • campbell_st
great 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
anonymous
  • anonymous
one sec

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