Obtain the equation of the parabola with focus (3,2) and directrix 3x-4y+9=0?

Question

1646 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T13:16:06

#16x^2+9y^2+24xy-204x-28y+244=0#

You suppose a point on the parabola #(x,y)#

The distance between the point #(x,y)# and the directrix is the same distance from the point #(x,y)# to the focus

#sqrt((x-3)^2+(y-2)^2)=|\3x-4y+9|/sqrt(3^2+(-4)^2)#

now by squaring both sides

#(x-3)^2+(y-2)^2=(3x-4y+9)^2/25#

Simplify

#25(x^2-6x+9+y^2-4y+4)=(3x+9)^2+(2)(-4y)(3x+9)+16y^2#

now multiply and simplify you get the following

#16x^2+9y^2+24xy-204x-28y+244=0#