Question

3x^2 + 7y^2 - 12x - 28y = -19. How do you find and what is the center?

Answer 1

bring it in the standard form first

Answer 2

@ishaan -g,-f is applicable for only circles.. this is not a circle.. the coefficients of x^2 and y^2 are different...

Answer 3

\[3(x^2-4x)+7(y^2-4y)=-19\]
\[3(x^2-4x+4)+7(y^2-4y+4)=-19+3(4)+7(4)\]
\[3(x-2)^2+7(y-2)^2=21\]
\[\frac{(x-2)^2}{7}+\frac{(y-2)^2}{3}=1\]

Answer 4

center of the ellipse is (2,2)

Answer 5

ah sorry i did that way too wrong