SOLUTION: What is the standard equation of: 4x^2 + 3y^2 + 8x - 24y + 51 = 0? I understand that it is an ellipse, and I have gotten to a point of: 4(x+1)^2 + 3(y-4)^2 = 1. However, I am un

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the standard equation of: 4x^2 + 3y^2 + 8x - 24y + 51 = 0? I understand that it is an ellipse, and I have gotten to a point of: 4(x+1)^2 + 3(y-4)^2 = 1. However, I am un      Log On


   



Question 814460: What is the standard equation of: 4x^2 + 3y^2 + 8x - 24y + 51 = 0?
I understand that it is an ellipse, and I have gotten to a point of: 4(x+1)^2 + 3(y-4)^2 = 1. However, I am unsure how to progress to get a^2 and b^2 to finish finding the equation.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What is the standard equation of: 4x^2 + 3y^2 + 8x - 24y + 51 = 0?
I understand that it is an ellipse, and I have gotten to a point of: 4(x+1)^2 + 3(y-4)^2 = 1. However, I am unsure how to progress to get a^2 and b^2 to finish finding the equation.
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4x^2+3y^2+8x-24y+51=0
4x^2+8x+3y^2-24y=-51
complete the square:
4(x^2+2x+1)+3(y^2-8y+16)=-51+4+48
4(x+1)^2+3(y-4)^2=1
%28x%2B1%29%5E2%2F%281%2F4%29%2B%28y-4%29%5E2%2F%281%2F3%29=1