What is the equation of the parabola whose vertex is ( 3 , 1 ) and directrix is 4 x + 3 y = 5 ?

1 Answer
Mar 12, 2018

Equation of parabola is #9x^2-24xy+16y^2-190x-80y+625=0#

Explanation:

Parabola is the locus of a point which moves so that its distance from a point, called focus, is equal to its distance from a given line, called directrix. Hence, if focus is #(h,k)# and directrix is #ax+by+c=0#, then equation of parabola is

#(x-h)^2+(y-k)^2=(ax+by+c)^2/(a^2+b^2)#

We know the directrix, but not the focus. However, vertex is exactly between focus and directrix. As equation of directrix is #4x+3y=5#, the equation of perpendicular from vertex to directrix is of the form #3x-4y=k# and as it passes through #(3,1)#, #k=5# and equation is #3x-4y=5#, which is line of symmetry.

Intersection of this line with directrix can be found by solving simultaneous equations #4x+3y=5# and #3x-4y=5# and it is #(7/5,-1/5)#. Let focus be #(x,y)#, then midpoint of #(x,y)# and #(7/5,-1/5)# is #(3,1)#. Hence #x=2*3-7/5=23/5# and #y=2*1+1/5=11/5#. Hence focus is #(23/5,11/5)#.

#(x-23/5)^2+(y-11/5)^2=(4x+3y-5)^2/(4^2+3^2)#

or #(5x-23)^2+(5y-11)^2=(4x+3y-5)^2#

or #25x^2-230x+529+25y^2-110y+121=16x^2+9y^2+25+24xy-40x-30y#

or #9x^2-24xy+16y^2-190x-80y+625=0#

graph{((x-23/5)^2+(y-11/5)^2-0.1)((x-3)^2+(y-1)^2-0.1)(9x^2-24xy+16y^2-190x-80y+625)(4x+3y-5)(3x-4y-5)=0 [-11.75, 28.25, -4.24, 15.76]}