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$$x^2+2xy+y^2+3x-3y-18=0$$

I don't know how to solve it.

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    what kind of variables are $$X,x,y$$? – 2017-01-29
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    Use $x^2+2xy+y^2=(x+y)^2$ – 2017-01-29
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    There are infinitely many pairs of solutions – 2017-01-29
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    x,y are real numbers – 2017-01-29
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    I did a very similar one about 15 minutes ago. Yours is a little easier, still a parabola http://math.stackexchange.com/questions/2119461/canonic-form-of-a-conic/2119687#2119687 – 2017-01-29
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    What do you mean by "solve" ? There infinitely many $(x,y)$ that satisfy the equation $(3,-3),(-6,0), (0,6)$... To be noticed: if $(x,y)$ is solution, $(y,-x)$ is solution. – 2017-01-29

4 Answers 4

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When a general conic equation

$$ Ax^2+Bxy+Cy^2+Dx +Ey +F=0$$

has $B\ne0$ and $A=C$ then rotating it $45^\circ$ will eliminate the $xy$ term.

This will be accomplished by the substitution

\begin{eqnarray} x&=&\frac{X-Y}{\sqrt{2}}\\ y&=&\frac{(X+Y)}{\sqrt{2}} \end{eqnarray}

\begin{equation} x^2+2xy+y^2+3x-3y-18=0 \end{equation} \begin{equation} \frac{(X-Y)^2}{2}+(X-Y)(X+Y)+\frac{(X+Y)^2}{2}+\frac{3}{\sqrt{2}}(X-Y)-\frac{3}{\sqrt{2}}(X+Y)-18=0 \end{equation}

$$X^2=\frac{3}{\sqrt{2}}Y+18$$ So it is the equation of a parabola rotated $45^\circ$ about the origin.

Parabola rotated 45 degrees

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    Question, is the parabola rotated about the origin or rotated about something else? – 2017-01-29
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    Yes, about the origin. – 2017-01-29
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    Hi,could you please tell me what background knowledge is required to solve it?I am lost on how did you solve it. @JohnWaylandBales – 2017-01-29
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    Here is a link to an explanation of the method used: https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.860401.html – 2017-01-29
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If you want to solve with respect to $x$ that write the equation as: $$ x^2+x(2y+3)+y^2-3y-18=0 $$ that use the quadratic formula $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ with $$ a=1 \qquad b=2y+3 \qquad c=y^2-3y-18 $$

If you want to solve with respect to $y$ do the same ordering the equation in $y$.

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Hint

$$ \color{red}{x^2}+\color{red}{2x}y+y^2+\color{red}{3x}-3y+18=\color{red}{\left(x+y+\frac{3}{2}\right)^2-3y-y^2-\frac{9}{4}}+y^2-3y-18 $$

Find a second square end solve the equation.

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You can use the transformation x=1/√2(a+b) y=1/√2(-a+b)