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

1 Answer
Jul 9, 2015

Answer:

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

Explanation:

There are various forms and formulas used for conic sections. I use #Ax^2 + Bxy +Cy^2 +Dx +Ey +F =0#

And #cot2theta = (A-C)/B# or #tan2theta = B/(A-C)#

So we re-write:

#−x+2y+x^2+xy+y^2=0# In the form:

#x^2 +xy +y^2 -x +2y =0#

We get: #cot2theta = (1-1)/1 = 0#,

so #2theta = pi/2 = 90^@#

and #theta = pi/4 = 45^@#

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

#x = hatxcostheta- haty sintheta#
#y = hatxsintheta+hatycostheta#

Using #theta = pi/4#, we get:

#x = hatx/sqrt2- haty/sqrt2#

#y = hatx/sqrt2+haty/sqrt2#

Replace and simplify.

I get

#(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.