Question 308967
{{{x^2  + 4y^2 - 2y = 8}}} Start with the given equation.



{{{x^2  + 4(y^2 - (1/2)y) = 8}}} Factor a 4 from the last two terms on the left side.



{{{x^2  + 4(y^2 - (1/2)y+1/16-1/16) = 8}}} Now take half of {{{1/2}}} to get {{{1/4}}}. Square it to get {{{1/16}}}. Add AND subtract this value inside the parenthesis.



{{{x^2  + 4((y^2 - (1/2)y+1/16)-1/16) = 8}}} Group up the first three terms inside the parenthesis.



{{{x^2  + 4((y-1/4)^2-1/16) = 8}}} Factor that inner group to get {{{(y-1/4)^2}}}



{{{x^2  + 4(y-1/4)^2-4(1/16) = 8}}} Distribute.



{{{x^2  + 4(y-1/4)^2-4/16 = 8}}} Multiply



{{{x^2  + 4(y-1/4)^2-1/4= 8}}} Reduce.



{{{x^2  + 4(y-1/4)^2= 8+1/4}}} Add {{{1/4}}} to both sides.



{{{x^2  + 4(y-1/4)^2= 33/4}}} Combine like terms.



{{{4x^2  + 16(y-1/4)^2= 33}}} Multiply EVERY term (outside the parenthesis) by the LCD 4 to clear out the fraction.



{{{(4x^2  + 16(y-1/4)^2)/33= 1}}} Divide both sides by 33 (to make the right side equal to 1).



{{{(4x^2)/33  + (16(y-1/4)^2)/33= 1}}} Break up the fraction.



{{{(x^2)/(33/4)  + ((y-1/4)^2)/(33/16)= 1}}} Rearrange the terms in the fractions.



{{{(x^2)/((sqrt(33)/2)^2)  + ((y-1/4)^2)/((sqrt(33)/4)^2)= 1}}} Rewrite {{{33/4}}} as {{{(sqrt(33)/2)^2}}} and {{{33/16}}} as {{{(sqrt(33)/4)^2}}}



{{{(x-0)^2/((sqrt(33)/2)^2)  + ((y-1/4)^2)/((sqrt(33)/4)^2)= 1}}} Finally, write {{{x^2}}} as {{{(x-0)^2}}}



It might be hard to notice, but the last equation above is in the form {{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} which is an ellipse.



So {{{(x-0)^2/((sqrt(33)/2)^2)  + ((y-1/4)^2)/((sqrt(33)/4)^2)= 1}}} is an ellipse



This means that {{{x^2  + 4y^2 - 2y = 8}}} is also an ellipse (since the two equations are equivalent)



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This is a lot of work to determine the conic. An alternative is to use the following rule:


For the general conic {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} 


If {{{B^2-4AC<0}}}, then the given conic above is an ellipse
Furthermore, if {{{A=C}}}, and {{{B=0}}}, then the conic is also a circle



If {{{B^2-4AC=0}}}, then the given conic above is a parabola



If {{{B^2-4AC>0}}}, then the given conic above is a hyperbola



In our case, we have the conic {{{x^2  + 4y^2 - 2y = 8}}} which means that {{{A=1}}}, {{{B=0}}}, {{{C=4}}}, {{{D=0}}}, {{{E=-2}}}, and {{{F=-8}}} (subtract this from both sides). Plug these values into {{{B^2-4AC}}} to get {{{0^2-4(1)(4)=-16}}} which is indeed less than 0 meaning that it is an ellipse.



Personally, I recommend doing it the long way for a while at first so you get a feel of what you are doing. Once you understand what's going on, take the shortcut.