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Question

# What is the standard form of the equation $x^2+6 y^2-2 x+12 y-23=0$?

Solution

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Arrange and complete the square to rewrite in standard form.

\begin{aligned} x^2+6y^2-2x+12y-23&=0\\ \left[x^2-2x\right]+\left[6y^2+12y\right]&=23&&\text{ APE}\\ \left[x^2-2x+\left(\frac{b}{2}\right)^2\right]+6\left[y^2+2y+\left(\frac{b}{2}\right)^2\right]&=23+\left(\frac{b}{2}\right)^2+6\left(\frac{b}{2}\right)^2&&\text{Complete the square}\\ \left[x^2-2x+\left(\frac{-2}{2}\right)^2\right]+6\left[y^2+2y+\left(\frac{2}{2}\right)^2\right]&=23+\left(\frac{-2}{2}\right)^2+6\left(\frac{2}{2}\right)^2&&\text{Substitute}\\ \left[x^2-2x+\left(-1\right)^2\right]+6\left[y^2+2y+\left(1\right)^2\right]&=23+\left(-1\right)^2+6\left(1\right)^2\\ \left(x^2-2x+1\right)6\left(y^2+2y+1\right)&=23+1+6\\ \left(x^2-2x+1\right)+6\left(y^2+2y+1\right)&=30&&\text{Factor}\\ (x-1)^2+6(y+1)^2&=30&&\text{(the equation should be equal to 1)}\\ \frac{(x-1)^2}{30}+\frac{6(y+1)^2}{30}&=\frac{30}{30}&&\text{ DPE}\\ \frac{(x-1)^2}{30}+\frac{(y+1)^2}{5}&=1 \end{aligned}

Hence, the standard form is $\dfrac{(x-1)^2}{30}+\dfrac{(y+1)^2}{5}=1$

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