Question 288363
{{{11x - x^2 - 4y^2 = 2y - 16}}} Start with the given equation.



{{{-x^2+11x-4y^2-2y=-16}}}  Subtract 2y from both sides and rearrange the terms.



{{{-(x-11/2)^2+121/4-4y^2-2y=-16}}} <a href="http://www.purplemath.com/modules/sqrquad.htm">Complete the square</a> for the x terms. Let me know if you need me to show you how to complete the square.



{{{-(x-11/2)^2+121/4-4(y+1/4)^2+1/4=-16}}} <a href="http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/completing-the-square.solver">Complete the square</a> for the y terms



{{{-(x-11/2)^2-4(y+1/4)^2+122/4=-16}}} Combine like terms



{{{-(x-11/2)^2-4(y+1/4)^2=-16-122/4}}}  Subtract {{{122/4}}} from both sides



{{{-(x-11/2)^2-4(y+1/4)^2=-186/4}}} Combine like terms



{{{-(x-11/2)^2-4(y+1/4)^2=-93/2}}} Reduce.



{{{(x-11/2)^2+4(y+1/4)^2=93/2}}} Multiply every term (outside the parenthesis) by -1.



{{{(2(x-11/2)^2)/93+(8(y+1/4)^2)/93=1}}} Multiply every term (outside the parenthesis) by {{{2/93}}} to make the right side equal to 1.



{{{((x-11/2)^2)/(93/2)+((y+1/4)^2)/(93/8)=1}}} Rewrite {{{2/93}}} as {{{1/(93/2)}}}. Rewrite {{{8/93}}} as {{{1/(93/8)}}}.



Now the equation is in the form {{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} where {{{h=11/2}}}, {{{k=-1/4}}}, {{{a^2=93/2}}} and {{{b^2=93/8}}} which is an equation of an ellipse.



So {{{((x-11/2)^2)/(93/2)+((y+1/4)^2)/(93/8)=1}}} is an ellipse.



Consequently, {{{11x-x^2-4y^2=2y-16}}} is also an ellipse (since the two equations are equivalent).