I can finish this if someone can help me get this equation into an equation of an ellipse like x^2/a^2 + y^2/b^2 = 1
The equation is: 16x^2 + 25y^2 - 64x - 100y + 384 = 0
I have worked it down to:
16(x-2)^2 + 25(y-2)^2 = -220
I'm pretty confident I'm headed in the right direction, but I can't figure out how to finish this. It would be helpful, if someone is daring enough, if I could also get the lengths of the major and minor axes, foci and eccentricity of this as well.
The equation is: 16x^2 + 25y^2 - 64x - 100y + 384 = 0
I have worked it down to:
16(x-2)^2 + 25(y-2)^2 = -220
I'm pretty confident I'm headed in the right direction, but I can't figure out how to finish this. It would be helpful, if someone is daring enough, if I could also get the lengths of the major and minor axes, foci and eccentricity of this as well.
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16(x² - 4x) + 25(y² - 4y) = -384
16(x² - 4x + 4) + 25(y² - 4y + 4) = -384 + 64 + 100
16(x - 2)² + 25(y - 2)² = -220
Yeah, that's what I got too. No idea how you can have a negative radius >.< Maybe it should be -384 instead of +384? They may have wanted it to be 400, so that you can divide out both multipliers in order to get the ellipse into the form:
x²/a² + y²/b² = 1
16(x² - 4x + 4) + 25(y² - 4y + 4) = -384 + 64 + 100
16(x - 2)² + 25(y - 2)² = -220
Yeah, that's what I got too. No idea how you can have a negative radius >.< Maybe it should be -384 instead of +384? They may have wanted it to be 400, so that you can divide out both multipliers in order to get the ellipse into the form:
x²/a² + y²/b² = 1
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Divide -220 to both sides to get the coefficient to be 1. Thats for the standard equation for an ellipse. x^2/a^2 + y^2/b^2 = 1 so the ellipse is horizontal.
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You did nothing wrong. Maybe the problem itself contains some error.