Using a directrix of #y=-2# and a focus of #(1, 6)#, what quadratic function is created?

1 Answer
Aniketh N.
Nov 23, 2016

#y= x^2/16-x/8+33/16#

Explanation:

#sqrt((x-1)^2+(y-6)^2)= sqrt((y+2)^2 #
Square both sides
#(x-1)^2+(y-6)^2=(y+2)^2#
subtract (y-6)^2 on both sides
#(x-1)^2=(y+2)^2-(y-6)^2#
Expand both sides
#x^2-2x+1=y^2+4y+4-(y^2-12y+36)#
Remove like terms on the right side
#x^2-2x+1=16y-32#
Add 32 to both sides
#x^2-2x+33=16y#
Divide both sides by 16
#x^2/16-(2x/16)+33/16=y#
You can reduce the 2x/16 to x/8 '
So your final answer is #x^2/16-x/8+33/16=y#

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