How to find the equation of a parabola whose focus is #(1,1)# and directrix is #x-y-3=0#?

1 Answer
Nov 30, 2017

Answer:

Equation of parabola is #x^2+2xy+y^2+2x-10y-5=0#

Explanation:

Parabola is the locus of a point which moves so that its distance from a given point called focus and its distance from a given line called directrix is always equal.

Let the point be #(x,y)#. Its distance from focus #(1,1)# is

#sqrt((x-1)^2+(y-1)^2)#

and its distance from a given line #x-y-3=0# is

#|(x-y-3)/(sqrt(1^2+1^2))|=|(x-y-3)/sqrt2|#

Hence equation of parabola is

#sqrt((x-1)^2+(y-1)^2)=|(x-y-3)/sqrt2|# and squaring

#(x-1)^2+(y-1)^2=((x-y-3)/sqrt2)^2=(x-y-3)^2/2#

or #2((x-1)^2+(y-1)^2)=(x-y-3)^2#

or #2x^2-4x+2+2y^2-4y+2=x^2+y^2+9-6x+6y-2xy#

or #x^2+2xy+y^2+2x-10y-5=0#

graph{(x^2+2xy+y^2+2x-10y-5)((x-1)^2+(y-1)^2-0.03)(x-y-3)=0 [-9.67, 10.33, -3.4, 6.6]}