find centre of ellipse
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find the center of the ellipse with the equation 3x^2 +4y^2+18x-32y-5=0
a.(3,-4) b.(-3,4) c.(4,-3) d.(-4,3)

You want to put the equation into standard form: (x - a)2/p2 + (y - b)2/q2 = 1. The centre is then (a, b).
To start, look at the two x-terms, 3x2 + 18x = 3(x2 + 6x). I want to complete the square so I add 9 insidethe parentheses. So to maintain the equality I also need to subtract 3 * 9 = 27 and hence I have
3x2 + 18x
= 3(x2 + 6x)
= 3(x2 + 6x + 9) - 27
= 3(x + 3)2 - 27.
Thus the equation of the ellipse can be written
3x2 + 4y2 + 18x - 32y - 5 = 0 which becomes
3(x + 3)2 -27 + 4y2 - 32y - 5 = 0 which becomes
3(x + 3)2 + 4y2 - 32y = 32.
Now perform the same technique with the two y-terms and then write the equation in standard form.
find the center of the ellipse with the equation 3x^2 +4y^2+18x-32y-5=0
a.(3,-4) b.(-3,4) c.(4,-3) d.(-4,3)

You want to put the equation into standard form: (x - a)2/p2 + (y - b)2/q2 = 1. The centre is then (a, b).
To start, look at the two x-terms, 3x2 + 18x = 3(x2 + 6x). I want to complete the square so I add 9 insidethe parentheses. So to maintain the equality I also need to subtract 3 * 9 = 27 and hence I have
3x2 + 18x
= 3(x2 + 6x)
= 3(x2 + 6x + 9) - 27
= 3(x + 3)2 - 27.
Thus the equation of the ellipse can be written
3x2 + 4y2 + 18x - 32y - 5 = 0 which becomes
3(x + 3)2 -27 + 4y2 - 32y - 5 = 0 which becomes
3(x + 3)2 + 4y2 - 32y = 32.
Now perform the same technique with the two y-terms and then write the equation in standard form.
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