How do you find the equation of the ellipse which satisfies the following conditions: the foci are at (+-2,0) and the ellipse passes through (2,-3)?

1 Answer
Apr 30, 2017

Answer:

Equation of ellipse is #x^2/16+y^2/12=1#

Explanation:

As the foci are #(+2,0)# and #(-2,0)#, the major axis is aligned with #x#-axis and centre of the ellipse is #(0,0)#. The equation of such an ellipse is of the type

#x^2/a^2+y^2/b^2=1#, where #2a# is major axis and #2b# is minor axis and #b^2=a^2(1-e^2)=a^2-a^2e^2#. where #e# is the eccentricity of the ellipse and foci are #(ae,0)# and #(-ae,0)#. Here #ae=2# and #a>2#.

Then the equation becomes #x^2/a^2+y^2/(a^2(1-e^2))=1#

As it passes through #(2,-3)#, we have

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

or #4(a^2-4)+9a^2=a^2(a^2-4)# and if #a^2=k# it becomes

#4k-16+9k=k^2-4k#

or #k^2-17k+16=0#

or #(k-1)(k-16)=0#

i.e. #k=a^2=1# or #k=a^2=16#

As #a>2#, we have #a=4# and #b=sqrt(4^2-2^2)sqrt12#

and equation of ellipse is #x^2/16+y^2/12=1#

graph{x^2/16+y^2/12=1 [-10, 10, -5, 5]}