What is the y coordinate for the center of the given ellipse?

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Find the y coordinate for the center of the ellipse given by the equation 8x^2 + y^2 - 4y = 2
 
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Try to rewrite the equation in the general form of an ellipse.

A (x - x0)2 + B (y - y0)2 = 0,

where the center is at x0, y0
 
i don't know how. how do you write it?
 
Do you know how to "complete the square"?
 
There is an easier method if you know differentiation. Differentiate the equation wrt x treating y as a constant to get a value for x, and similarly for y by treating x as a constant. The values of x and y you get are the coordinates of the center.
 
the center is at (0,3)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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