anonymous
  • anonymous
Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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campbell_st
  • campbell_st
have you any method to use for this question...?
anonymous
  • anonymous
not really no
campbell_st
  • campbell_st
so what methods do you know.. 1. find the focal length and then find the vertex... 2. use the distance formula..

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anonymous
  • anonymous
number 1
campbell_st
  • campbell_st
ok.... so start by plotting the given information |dw:1437680444218:dw| so this tells me the parabola is concave up... does that make sense...
anonymous
  • anonymous
yes
campbell_st
  • campbell_st
ok... so the line of symmetry is x = -5 and the vertex is a units below the focus... where a = the focal length so |dw:1437680607256:dw|
anonymous
  • anonymous
so the length between the focus and the directrix is 6
campbell_st
  • campbell_st
the distance between the focus and directrix is 2a the vertex is midway between the vertex and directrix on the line of symmetry
campbell_st
  • campbell_st
that's great so 2a = 6 therefore the focal length is a = 3 is that ok..?
anonymous
  • anonymous
yes
campbell_st
  • campbell_st
|dw:1437680777548:dw|
campbell_st
  • campbell_st
so where do you think the vertex is located..?
anonymous
  • anonymous
woud it be at (-5,2)
campbell_st
  • campbell_st
perfect.... now to get the correct equation I use a standard form \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length so the equation is \[(x + 5)^2 = 4 \times 3 (y - 2)\] so this can be simplified and rewritten in the form you need.
campbell_st
  • campbell_st
hope it all makes sense
anonymous
  • anonymous
Thank you, it really does help
campbell_st
  • campbell_st
if you need it rewritten I'd use \[y = \frac{1}{12}(x + 5)^2 + 2\]
anonymous
  • anonymous
That's exactly what I thought it was. Thank you so much

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