How do you find center, vertex, and foci of an ellipse #9x^2 + 25y^2 - 36x + 50y - 164 = 0#?

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Answer 1 · 2016-02-15T09:54:59

Center of ellipse is #(2,-1)#, vertices are #(-3, -1)# and #(7, -1)# and foci are #(-2, -1)# and #(6, -1)#.

The equation #9x^2+25y^2-36x+50y-164=0# can be written as

#9(x^2-4x)+25(y^2+2y)=164# or

#9(x^2-4x+4)+25(y^2+2y+1)=164+36+25#

#9(x-2)^2+25(y+1)^2=225#

Dividing both sides 225, we get

#9/225(x-2)^2+25/225(y+1)^2=1# or #(x-2)^2/5^2+(y+1)^2/3^2=1#

Hence, Center of ellipse is #(2,-1)# and major axis is #10# (#2*5#) and minor axis is #6# (#2*3#) and vertices are #(-3, -1)# and #(7, -1)#.

To find foci, #c^2=5^2-3^2=16=4^2#. Hence foci are #(-2, -1)# and #(6, -1)# (4 units on major axis on either side of center).