anonymous
  • anonymous
find the equation of the parabola given the vertex (4,6) and focus (4,0)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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tkhunny
  • tkhunny
What are your plans?
anonymous
  • anonymous
would (4,6) be the center so its in the form (x-4)^2=4p(y-6)
triciaal
  • triciaal
|dw:1446443265228:dw|

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anonymous
  • anonymous
|dw:1446436326567:dw|
triciaal
  • triciaal
|dw:1446443399139:dw|
anonymous
  • anonymous
how do u know it passes through (0,0) and (8,0)
triciaal
  • triciaal
we need to find out the x coordinate for the vertex is the line of symmetry x = 4 the y value is the f(the x value) standard ax^2 + bx + c vertex form a(x-h)^2 + k h = -b/2a
anonymous
  • anonymous
ok
anonymous
  • anonymous
is the x value =4?
triciaal
  • triciaal
yes
anonymous
  • anonymous
so y value = f(4) ?
triciaal
  • triciaal
the vertex is (4, 6) the focus has to be "inside" so it opens down the directrix is the horizontal on the opposite side of the focus same distance from the vertex
anonymous
  • anonymous
ok
anonymous
  • anonymous
the directrix would be (4,8) or the line y=8
anonymous
  • anonymous
right?
anonymous
  • anonymous
or is it (4,12) so y=12
anonymous
  • anonymous
or is it (4,12) so y=12
campbell_st
  • campbell_st
here is a really simple method, after graphing the points you can see the parabola is concave down so find the perpendicular distance between the vertex and focus...its called the focal length and because of the concavity, the focal length will be a negative then use the general form \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length... substitute them and you have the equation. if necessary make y the subject hope it helps
triciaal
  • triciaal
|dw:1446444768446:dw|
campbell_st
  • campbell_st
if you find the focal length... it avoids all the other complicated stuff...
anonymous
  • anonymous
ok
campbell_st
  • campbell_st
so how far from the vertex to the focus.. ?
anonymous
  • anonymous
6 units
campbell_st
  • campbell_st
great so are you happy the parabola is concave down..?
anonymous
  • anonymous
yes
triciaal
  • triciaal
|dw:1446445025276:dw|
campbell_st
  • campbell_st
ok... so a = -6 and the vertex is (4, 6) so substitute them into \[(x - h)^2 = 4 \times a (y - k)\] remember (h, k) is the vertex... what do you get
anonymous
  • anonymous
\[(x-4)^2=-24(y-6)\]
campbell_st
  • campbell_st
great... this is a vertex form.... but you most likely want \[y = p(x - h)^2 + k\] so from the equation you have entered, can you make y the subject...?
anonymous
  • anonymous
ok but thats the form it have to be in ..but we can solve it the other way too
anonymous
  • anonymous
\[-\frac{ 1 }{ 24 }(x-4)^2+6=y\]
campbell_st
  • campbell_st
you could solve it... the value of a is the focal length... the directrix is a units about the vertex... so would have an equation y = 12 you can then use the distance formula to find the equation... but for me, finding the focal length and then using a vertex form, I'd expect. you haven't seem is so much easier in the end if you have \[(x - 4)^2 = -24(y - k)\] divide both sides by -24 \[y - 6 = \frac{-1}{24} ( x-4)^2 \] the you can see the rest... I don't know why you're not taught this method.... but I do hope it makes some sense...
campbell_st
  • campbell_st
sorry to butt in @triciaal , but I thought it would avoid confusion by just plotting the points as you did then find the focal length.
anonymous
  • anonymous
thanks to you both
triciaal
  • triciaal
@campbell_st no problem just didn't know what I did wrong @AliLnn now you have 2 methods you are welcome

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