How do I graph the ellipse with the equation $- x + 2 y + x^{2} + x y + y^{2} = 0$ $−x+2y+x^2+xy+y^2=0$ ?

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Answer 1 · 2015-07-09T15:50:15

Remove the cross-product term, then graph on the new coordinate axes.

There are various forms and formulas used for conic sections. I use $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ $Ax^2 + Bxy +Cy^2 +Dx +Ey +F =0$

And $cot 2 θ = \frac{A - C}{B}$ $cot2theta = (A-C)/B$ or $tan 2 θ = \frac{B}{A - C}$ $tan2theta = B/(A-C)$

So we re-write:

$- x + 2 y + x^{2} + x y + y^{2} = 0$ $−x+2y+x^2+xy+y^2=0$ In the form:

$x^{2} + x y + y^{2} - x + 2 y = 0$ $x^2 +xy +y^2 -x +2y =0$

We get: $cot 2 θ = \frac{1 - 1}{1} = 0$ $cot2theta = (1-1)/1 = 0$ ,

so $2 θ = \frac{π}{2} = 90^{\circ}$ $2theta = pi/2 = 90^@$

and $θ = \frac{π}{4} = 45^{\circ}$ $theta = pi/4 = 45^@$

Our new coordinate system will be denoted $ˆ x$ $hat(x)$ and $ˆ y$ $hat(y)$ . To get this equations replace $x$ $x$ and $y$ $y$ by:

$x = ˆ x cos θ - ˆ y sin θ$ $x = hatxcostheta- haty sintheta$
$y = ˆ x sin θ + ˆ y cos θ$ $y = hatxsintheta+hatycostheta$

Using $θ = \frac{π}{4}$ $theta = pi/4$ , we get:

$x = \frac{ˆ x}{\sqrt{2}} - \frac{ˆ y}{\sqrt{2}}$ $x = hatx/sqrt2- haty/sqrt2$

$y = \frac{ˆ x}{\sqrt{2}} + \frac{ˆ y}{\sqrt{2}}$ $y = hatx/sqrt2+haty/sqrt2$

Replace and simplify.

I get

$\frac{3 ˆ x}{2} + \frac{ˆ x}{\sqrt{2}} + \frac{{ˆ y}^{2}}{2} + \frac{3 ˆ y}{\sqrt{2}} = 0$ $(3hatx)/2 + hatx/sqrt2 + haty^2/2 +(3haty)/sqrt2 = 0$

Now treat it like a non-rotated ellipse. Find center and vertices and graph on the new axes.

How do I graph the ellipse with the equation −x+2y+x^2+xy+y^2=0−x+2y+x2+xy+y2=0−x+2y+x^2+xy+y^2=0?