The tangent at p(3,6) on the parabola y²=12x meets the tangent at the vertex at K. Prove that the line through K parallel to the normal at P passes through the focus of the parabola?

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Answer 1 · 2018-05-16T09:24:15

Please see below.

Focus of a parabola whose equation is #y^2=4ax# is #(a,0)# and hence focus of parabola #y^2=12x# is #(3,0)#.

Slope of tangent is given by first derivative #y^2=12x# i.e.

#2y(dy)/(dx)=12# and #(dy)/(dx)=12/(2y)=6/y#

Hence slope of tangent at #(3,6)# is #6/6=1# and its equation is

#y-6=1(x-3)# or #x-y+3=0#

As normal is perpendicular to tangent, its slope is #-1/1=-1# and equation is

#y-6=-1(x-3)# or #x+y=9#

As #y^2=12x# in vertex form is #x=1/12(y-0)^2+0#, its vertex is #(0,0)# and as parabola is horizontal, tangent at vertex is #x=0#

Point of intersection of #x-y+3=0# and #x=0# can be obtained by solving these simultaneous equations, which gives coordinates of #K# as #(0,3)#.

Now as line through #K(0,3)# parallel to the normal at #P# will also have a slope of #-1#, its equation is

#y-3=-x# or #x+y-3=0#

As coordinates of focus #(3,0)# satisfy the equation #x+y-3=0#, focus lies on this.

graph{(y^2-12x)(x-y+3)(x+y-9)(x+y-3)((x-3)^2+(y-6)^2-0.02)((x-3)^2+y^2-0.02)=0 [-8.5, 11.5, -2.24, 7.76]}