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

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted July 16, 2013 at 11:30 PM via web

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Given the ellipse 8x^2 + y^2 + 80x - 6y + 193 = 0 find:

a) The Center C

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lfryerda | High School Teacher | (Level 2) Educator

Posted July 16, 2013 at 11:41 PM (Answer #1)

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If you have more than one question, you need to make separate posts.

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

`(x+5)^2/2+(y-3)^2/16=1`

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

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