Given the ellipse 8x^2 + y^2 + 80x - 6y + 193 = 0 find: a) The Center C



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Posted on (Answer #1)

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To find the center of an ellipse, it needs to be put into standard form. This means that you need complete the square for both the x and y coordinates.

`8x^2+y^2+80x-6y+193=0`  collect terms for x and y

`8x^2+80x+y^2-6y+193=0`   factor 8 from x terms

`8(x^2+10x)+y^2-6y+193=0`  add and subtract to complete the square

`8(x^2+10x+25-25)+y^2-6y+9-9+193=0`  arrange perfect squares

`8(x+5)^2-8(25)+(y-3)^2-9+193=0`  collect constants

`8(x+5)^2+(y-3)^2=200+9-193`  simplify

`8(x+5)^2+(y-3)^2=16`   divide by 16


The centre of the ellipse is at (-5,3).

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