What is the equation of a parabola with the focus located at (2,5) and the directrix y=1

Question

What is the equation of a parabola with the focus located at (2,5) and the directrix y=1

 

Answer 1

For the directrix and focus of a parabola, the distance from any point (x,y) on the parabola to the focus

is the same as the straight line distance from that point to the directrix.  So we use the distance formula to

come up with equations for the two distances and set those equal to each other.

 

For focus at (a,b) and directrix y=c

√((x-a)²+(y-b)²) = absolute value [y-c]

 

We have a=2, b=5, c=1 from information given so

 

√((x-2)²+(y-5)²) =absolute value[y-1]

Square both sides

(x-2)² + (y-5)² = (y-1)²

x²-4x+4+y²-10y+25 = y²-2y+1

 

Simplify by adding like terms and bringing all terms to one side

 

x²-4x+y²-y²-10y+2y+29-1 =0

x²-4x-8y+28 = 0

Let's move the y term back to the right side

 

8y = x²-4x+28

Divide both sides by 8

 

y = (1/8)x² - (1/2)x + 7/2