What is the equation for a parabola with vertex:(8,6) and focus:(3,6)?

1 Answer
Jul 19, 2016

For the parabola it is given

#V->"Vertex"=(8,6)#

#F->"Focus"=(3,6)#

We are to find out the equation of the parabola

The ordinates of V(8,6) and F(3,6) being 6 the axis of parabola will be parallel to x-axis and its equation is #y=6#

Now let the coordinate of the point (M) of intersection of directrix and axis of parabola be #(x_1,6)#.Then V will be mid point of MF by the property of parabola. So

#(x_1+3)/2=8=>x_1=13#
#"Hence"M->(13,6)#

The directrix which is perpendicular to the axis (#y=6#) will have equation #x=13 or x-13=0#

Now if# P(h,k)# be any point on the parabola and N is the foot of the perpendicular drawn from P to the directrix,then by the property of parabola

#FP=PN#

#=>sqrt((h-3)^2+(k-6)^2)=h-13#

#=>(h-3)^2+(k-6)^2=(h-13)^2#

#=>(k-6)^2=(h-13)^2-(h-3)^2#

#=>(k^2-12k+36=(h-13+h-3)(h-13-h+3)#

#=>k^2-12k+36=(2h-16)(-10)#

#=>k^2-12k+36+20h-160=0#

#=>k^2-12k+20h-124=0#

Replacing h by x and k by y we get the required equation of the parabola as

#color(red)(y^2-12y+20x-124=0)#