Find the equation of an ellipse centered at origin and passing through point #(-3,2)# whose latus rectum is #3# times the distance from center to focus?

1 Answer
Nov 15, 2017

Equation of ellipse is #x^2/12+y^2/16=1#, if major axis is along #y#-axis and #3x^2+4y^2=43# if major axis is along #x#-axis.

Explanation:

As the major axis and minor axis coincide with #y#-axis and #x#-axis, the center of ellipse is #(0,0)# and its equation is of the form

#x^2/a^2+y^2/b^2=1#

In such an ellipse, focus is at #(+-ae,0)# (if majoraxis is along #x#-axis) or #(0,+-ae)# (if majoraxis is along #y#-axis).

Let us assume it is along #y#-axis and then distance between center and focus is #be# and #e# the eccentricity of ellipse given by #e=sqrt(1-a^2/b^2)# and length of latus rectum is #(2a^2)/b=2b(1-e^2)#.

As latus rectum is #3# times the distance from center to focus, we should have

#2b(1-e^2)=3be# or #2-2e^2=3e# or #2e^2+3e-2=0#

or #(e+2)(2e-1)-0#

But as #e# is positive, we have #e=1/2# and

#1-a^2/b^2=1/4# or #a^2/b^2=3/4# i.e. #a^2=3/4b^2#

Hence equation of ellipse is #(4x^2)/(3b^2)+y^2/b^2=1#

Now it passes through #(-3,2)#, hence

#(4xx9)/(3xxb^2)+4/b^2=1#

or #b^2=12+4=16#

and equation of ellipse is #(4x^2)/48+y^2/16=1#

or #x^2/12+y^2/16=1#

graph{4x^2+3y^2=48 [-10, 10, -5, 5]}

If major axis is along #x#-axis and then distance between center and focus is #ae#. Here #e# is the eccentricity of ellipse given by #e=sqrt(1-b^2/a^2)# and length of latus rectum is #(2b^2)/a=2a(1-e^2)#.

As latus rectum is #3# times the distance from center to focus, we should have

#2a(1-e^2)=3ae# or #2-2e^2=3e# or #2e^2+3e-2=0#

or #(e+2)(2e-1)-0#

But as #e# is positive, we have #e=1/2# and

#1-b^2/a^2=1/4# or #b^2/a^2=3/4# i.e. #b^2=3/4a^2#

Hence equation of ellipse is #x^2/a^2+(4y^2)/(3a^2)=1#

Now it passes through #(-3,2)#, hence

#9/a^2+(4xx4)/(3a^2)=1#

or #a^2=9+16/3=43/3#

and equation of ellipse is #(3x^2)/43+(4y^2)/43=1#

or #3x^2+4y^2=43#

graph{3x^2+4y^2=43 [-8, 8, -4, 4]}