Q . If the focus of a parabola is ( -1 , 1) and directrix is 4x + 3 y - 24 =0 then find its axis vertex and length of Latus rectum.?

1 Answer
Mar 21, 2018

Axis of symmetry is #3x-4y+7=0#, vertex is #(1,5/2)# and length of Latus rectum is #10#.

Explanation:

As axis of symmetry is perpendicular to directrix #4x+3y-24=0#, equation of axis of symmetry is of the form #3x-4y=k# and as it passes through focus #(-1,1)#, #k=3*(-1)-4*1=-7#.

Hence axis of symmetry is #3x-4y+7=0#

Foot of perpendicular from focus on to dirctrix is point of intersection of #4x+3y-24=0# and #3x-4y+7=0# i.e. #(3,4)#. Andas vertex is midway between this point and focus, vertex is #((3-1)/2,(4+1)/2)# i.e. #(1,5/2)#. Length of perpendicular from focus to directrix is #|(4(-1)+3*1-24)/sqrt(4^2+3^2)|=5# and hence length of Latus rectum is #10#.

Now, parabola is the locus of a point #(x,y)#, which moves so that its distance from focus #(-1,1)# and directrix #4x+3y-24=0# are equal, hence equation of parabola is

#(x+1)^2+(y-1)^2=|(4x+3y-24)/sqrt(4^2+3^2)|^2#

or #25(x+1)^2+25(y-1)^2=(4x+3y-24)^2#

or #25x^2+50x+25+25y^2-50y+25=16x^2+9y^2+576-192x-144y+24xy#

or #9x^2+16y^2-24xy+242x+94y-526=0#

graph{(9x^2+16y^2-24xy+242x+94y-526)((x+1)^2+(y-1)^2(x+1)^2+(y-1)^2-0.03)(4x+3y+1)(4x+3y-24)(3x-4y+7)=0 [-10.67, 9.33, -3.68, 6.32]}