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What is the standard form of the equation x2+6y22x+12y23=0x^2+6 y^2-2 x+12 y-23=0?

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Answered 2 years ago
Answered 2 years ago
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Arrange and complete the square to rewrite in standard form.

x2+6y22x+12y23=0[x22x]+[6y2+12y]=23 APE[x22x+(b2)2]+6[y2+2y+(b2)2]=23+(b2)2+6(b2)2Complete the square[x22x+(22)2]+6[y2+2y+(22)2]=23+(22)2+6(22)2Substitute[x22x+(1)2]+6[y2+2y+(1)2]=23+(1)2+6(1)2(x22x+1)6(y2+2y+1)=23+1+6(x22x+1)+6(y2+2y+1)=30Factor(x1)2+6(y+1)2=30(the equation should be equal to 1)(x1)230+6(y+1)230=3030 DPE(x1)230+(y+1)25=1\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 (x1)230+(y+1)25=1\dfrac{(x-1)^2}{30}+\dfrac{(y+1)^2}{5}=1

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