anonymous
  • anonymous
Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.
Mathematics
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anonymous
  • anonymous
Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.
Mathematics
katieb
  • katieb
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campbell_st
  • campbell_st
ok... have you plotted the information on the number plane so you know the concavity of the parabola..?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
So, you can still use the combined distance equation for this problem. But you just have to make your solution ax^2+bx+c=0.

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anonymous
  • anonymous
ax+by=c.
campbell_st
  • campbell_st
so is the directrix above or below the focus...? and then is the parabola concave up or down..?
campbell_st
  • campbell_st
|dw:1438892282167:dw| so you plot looks like this
anonymous
  • anonymous
below right?
campbell_st
  • campbell_st
|dw:1438892378795:dw| the distance is labelled 2a and is double the focal length.
anonymous
  • anonymous
8
campbell_st
  • campbell_st
the dotted line through x = 4 is the line of symmetry.... and the vertex is on this line midway between the focus and directrix
campbell_st
  • campbell_st
great so if the twice the focal length is 8 then the focal length is a = 4 so the vertex is 4 units below the focus on the line x = 4 so where do you think the vertex is..?
campbell_st
  • campbell_st
|dw:1438892601545:dw|
anonymous
  • anonymous
at y=-11 ??
anonymous
  • anonymous
or no
campbell_st
  • campbell_st
great... so the standard form I use now is \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length so the equation is \[(x -4)^2 = 4 \times 4(y + 11)\] now you just need to simplify this equation
anonymous
  • anonymous
f(x)=1/16x^-8x+11 ?
campbell_st
  • campbell_st
so if you want it in vertex form its \[\frac{1}{16}(x -4)^2 = y + 11\] then subtract 11 from both sides
campbell_st
  • campbell_st
well not quite \[\frac{1}{16}(x^2 - 8x + 16)- 11 = y\]
campbell_st
  • campbell_st
which would become \[y = \frac{x^2}{16} - \frac{x}{2} - 10 \]
anonymous
  • anonymous
ohh ok, thats the same as 1/16x^2-1/2x-10 right?
campbell_st
  • campbell_st
it is the same
anonymous
  • anonymous
ok thank you so much :)

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