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Question 1204031: Suppose a parabola opens upward, has focus at (1,-1), has a horizontal directrix, and passes through the point (-31,59).
a. The directrix will have an equation of?
b. The equation of the parabola will be?
Found 2 solutions by math_tutor2020, MathLover1:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
F = focus = (1,-1)
P = point on parabola = (-31,59)
D = point on directrix that is directly above or below P (this is for a horizontal directrix only)
For a parabola to be possible, we must have this condition hold:
FP = PD
Check out the diagram shown here
Image Source:
https://en.wikipedia.org/wiki/Parabola
The pink congruent segments are what you should focus on.
Use the distance formula to see how far it is from the focus F(1,-1) to the point on the parabola P(-31,59)
(x1,y1) = (1,-1) and (x2,y2) = (-31,59)
The distance from the focus F(1,-1) to a point on the parabola P(-31,59) is exactly 68 units.
This is the length of segment FP.
It's also the length of segment PD.
Due to the horizontal directrix, from point P(-31,59), we will either move up 68 units or move down 68 units to arrive at point D.
Because P has a y coordinate larger than the y coordinate of F, it must mean the parabola opens upward. Consequently it means the directrix is below point P.
So we'll move down 68 units from P(-31,59) to D(-31,-9)
The equation of the directrix is then the y coordinate of point D.
Answer: The equation of the directrix is y = -9
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Part (b)
Start at the focus F(1,-1) and move straight down to arrive at the directrix y = -9
This will move you 8 units down.
Half of which is 4.
This is the focal length. It's the distance between vertex and focus. It's also the distance from vertex to directrix.
Start at F(1,-1) to move 4 units down to arrive at the vertex V(1,-5)
Recall the vertex form template is
y = a(x-h)^2+k
where
a = determines whether the parabola opens up or down
(h,k) = vertex
Plug in the coordinates of the vertex to get
y = a(x-h)^2+k
y = a(x-1)^2+(-5)
y = a(x-1)^2-5
Then plug in the coordinates of the other point on the parabola (-31,59). This will allow us to solve for 'a'
y = a(x-1)^2-5
59 = a(-31-1)^2-5
59 = a(-32)^2-5
59 = 1024a-5
1024a-5 = 59
1024a = 59+5
1024a = 64
a = 64/1024
a = (1*64)/(16*64)
a = 1/16
a = 0.0625 exactly
Therefore the vertex form equation would be
y = (1/16)(x-1)^2-5
or
y = 0.0625(x-1)^2-5
I'll expand out the decimal form
y = 0.0625(x-1)^2-5
y = 0.0625(x^2-2x+1)-5
y = 0.0625x^2-0.125x+0.0625-5
y = 0.0625x^2-0.125x-4.9375
This is in standard form y = ax^2+bx+c
a = 0.0625 = 1/16
b = -0.125 = -1/8
c = -4.9375 = -79/16
Each decimal is exact and hasn't been rounded.
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Answer
The equation of the parabola in standard form, using fractions, is y = (1/16)x^2 - (1/8)x - 79/16
That converts to the exact decimal form y = 0.0625x^2-0.125x-4.9375
Or to save time, you can stick to the vertex form y = (1/16)(x-1)^2-5 aka y = 0.0625(x-1)^2-5
It will depend on which form your teacher will want.
Answer by MathLover1(20850)  (Show Source):
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