What is the equation of the parabola passing through the points (0, 0) and (0,1) and having the line x+y+1=0 as its axis of symmetry?

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Answer 1 · 2017-01-25T09:57:43

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

As axis of symmetry is #x+y+1=0# and focus lies on it, if abscissa of focus is #p#, ordinate is #-(p+1)# and coordinates of focus are #(p,-(p+1))#.

Further, directrix will be perpendicular to axis of symmetry and its equation would be of the form #x-y+k=0#

As every point on parabola is equidistant from focus and directrix, its equation will be

#(x-p)^2+(y+p+1)^2=(x-y+k)^2/2#

This parabola passes through #(0,0)# and #(0,1)# and hence

#p^2+(p+1)^2=k^2/2# .....................(1) and

#p^2+(p+2)^2=(k-1)^2/2# .....................(2)

Subtracting (1) from (2), we get

#2p+3=(-2k+1)/2#, which gives #k=-2p-5/2#

This reduces equation of parabola to #(x-p)^2+(y+p+1)^2=(x-y-2p-5/2)^2/2#

and as it passes through #(0,0)#, we get

#p^2+p^2+2p+1=(4p^2+10p+25/4)/2# or #4p+2=25/4+10p#

i.e. #6p=-17/4# and #p=-17/24#

and hence #k=-2xx(-17/24)-5/2=-13/12#

and equation of parabola as

#(x+17/24)^2+(y+7/24)^2=(x-y-13/12)^2/2# and multiplying by #576=24^2#, we get

or #(24x+17)^2+(24y+7)^2=2(12x-12y-13)^2#

or #576x^2+816x+289+576y^2+336y+49=2(144x^2+144y^2+169-288xy-312x+312y#

or #288x^2+288y^2+576xy+1440x-288y=0#

or #x^2+y^2+2xy+5x-y=0#
graph{(x^2+y^2+2xy+5x-y)(x+y+1)(12x-12y-13)=0 [-11.42, 8.58, -2.48, 7.52]}