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What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix?

 Jun 15, 2017

Best Answer 

 #2
avatar+2440 
+1

There's an excellent formula that I concocted 5 days ago to always give the equation of a parabola when only the directrix and focus is given. Here is the equation:

 

Let a = x-coordinate of focus

Let b = y-coordinate of focus

Let k= equation of line of the directrix

\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\)

 

All you have to do is plug into this formula, simplify, and you're done. Let's try this together:

 

a=-2

b=4

k=6
 

\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\) Plug in the appropriate values that are given by the focus and directrix.
\(y=\frac{(x-(-2))^2}{2(4-6)}+\frac{4+6}{2}\) Let's clean this up a bit, shall we?
\(y=\frac{(x+2)^2}{-4}+5\) Technicaly, you could stop here and call it a day, but I am going to attempt to make it look even cleaner! I'll expand the \((x+2)^2\)
\(y=-\frac{x^2+4x+4}{4}+5\) After expanding it, I'll change 5 into an improper fraction that I can add to the current fraction. 
\(y=-\frac{x^2+4x+4}{4}+\frac{20}{4}\) Since the fractions have common denominators. Add the fractions together, but you have to be very attentive to how you do this. Notice how there is a negative. I'm going to get rid of this because you can't combine a negative fraction with a positive one; it just doesn't work.
\(y=\frac{-(x^2+4x+4)}{4}+\frac{20}{4}\) I'm distributing that negative because you cannot combine a negative and positive fraction.
\(y=\frac{-x^2-4x-4}{4}+\frac{20}{4}\) Now, you can add the fractions together. Now you add the fractions together normally.
\(y=\frac{-x^2-4x+16}{4}\) Break off the two last terms from the current fraction. You'll see why.
\(y=\frac{-x^2}{4}+\frac{-4x+16}{4}\) The rightmost fraction can be simplified because a factor of 4 goes into both terms. Wow!
\(y=\frac{-x^2}{4}-x+4\)  
\(y=-\frac{1}{4}x^2-x+4\) This is your final equation.
   

 

No, I did not just pull that formula out of the air! I derived it myself. If you are wondering why this formula will work 100% of the time, I explained it, in detail, here where someone asked a similar question to yours. 

 

https://web2.0calc.com/questions/parabola-equation#r3

 Jun 15, 2017
 #1
avatar+128407 
+1

 

We want the form

 

4p (y - k)  =  (x - h)^2

 

Where  the vertex is (-2 ,6+ p)    and  p = ( y coordinate of the focus - y value of the directrix) / 2 =( 4 - 6)/2  = -2/2  = -1

 

So the vertex  = (-2, 5)  = (h,k)

 

So we have

 

4(-1) (y -5)  = ( x - -2)^2    simplify

 

-4(y - 5) = ( x + 2)^2

 

-4y + 20  =  x^2 + 4x + 4   subtract 20 from both sides

 

-4y  = x^2 + 4x - 16          divide through  by  -4

 

y = (-1/4)x^2  - x  + 4

 

Here's the graph : https://www.desmos.com/calculator/1khnfskw3p

 

 

 

cool cool cool

 Jun 15, 2017
 #2
avatar+2440 
+1
Best Answer

There's an excellent formula that I concocted 5 days ago to always give the equation of a parabola when only the directrix and focus is given. Here is the equation:

 

Let a = x-coordinate of focus

Let b = y-coordinate of focus

Let k= equation of line of the directrix

\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\)

 

All you have to do is plug into this formula, simplify, and you're done. Let's try this together:

 

a=-2

b=4

k=6
 

\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\) Plug in the appropriate values that are given by the focus and directrix.
\(y=\frac{(x-(-2))^2}{2(4-6)}+\frac{4+6}{2}\) Let's clean this up a bit, shall we?
\(y=\frac{(x+2)^2}{-4}+5\) Technicaly, you could stop here and call it a day, but I am going to attempt to make it look even cleaner! I'll expand the \((x+2)^2\)
\(y=-\frac{x^2+4x+4}{4}+5\) After expanding it, I'll change 5 into an improper fraction that I can add to the current fraction. 
\(y=-\frac{x^2+4x+4}{4}+\frac{20}{4}\) Since the fractions have common denominators. Add the fractions together, but you have to be very attentive to how you do this. Notice how there is a negative. I'm going to get rid of this because you can't combine a negative fraction with a positive one; it just doesn't work.
\(y=\frac{-(x^2+4x+4)}{4}+\frac{20}{4}\) I'm distributing that negative because you cannot combine a negative and positive fraction.
\(y=\frac{-x^2-4x-4}{4}+\frac{20}{4}\) Now, you can add the fractions together. Now you add the fractions together normally.
\(y=\frac{-x^2-4x+16}{4}\) Break off the two last terms from the current fraction. You'll see why.
\(y=\frac{-x^2}{4}+\frac{-4x+16}{4}\) The rightmost fraction can be simplified because a factor of 4 goes into both terms. Wow!
\(y=\frac{-x^2}{4}-x+4\)  
\(y=-\frac{1}{4}x^2-x+4\) This is your final equation.
   

 

No, I did not just pull that formula out of the air! I derived it myself. If you are wondering why this formula will work 100% of the time, I explained it, in detail, here where someone asked a similar question to yours. 

 

https://web2.0calc.com/questions/parabola-equation#r3

TheXSquaredFactor Jun 15, 2017

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