Help! Rotate the axis to eliminate the xy-term in the equation then write the equation ins standard form. x^2-4xy+4y^2+5sqrt(5)+1=0 Help!?

3 Answers
Dec 3, 2016

Please see below.

Explanation:

The equation given here #x^2-4xy+4y^2+5sqrt5+1=0# is of the form #Ax^2+Bxy+Cy^2+Dx+Ey+F=0#.

What a rotation does is it changes #x# & #y#-axes to #x'# & #y'#-axes, as shown below,.
enter image source here
In such a case, the relation between coordinate #(x,y)# and new coordinates #(x',y')# is given by

#x=x'costheta-y'sintheta# and #y=x'sintheta+y'costheta#

and reverse is #x'=xcostheta+ysintheta# and #y'=-xsintheta-costheta#

Note that latter equations are equivalent to rotation by #-theta#.

In such cases, we can eliminate #xy# if rotated by #theta=(A-C)/B#

In given equation we have #cot2theta=(1-4)/(-4)=3/4# i.e.

#(cot^2theta-1)/(2cottheta)=3/4# or #4cot^2theta-6cottheta-4=0#

or #(2cottheta-4)(2cottheta+1)=0# i.e. #cottheta=2# or #-1/2#

These two angles relate to #theta# and #theta-90^o# in the image above. Working out for #cottheta=2#

Hence, either #sintheta=1/sqrt5# and #costheta=2/sqrt5#

or #sintheta=-2/sqrt5# and #costheta=1/sqrt5#

and we have #x=(2x')/sqrt5+(y')/sqrt5# and #y=(x')/sqrt5+(2y')/sqrt5#

and putting these in given equation and simplifying we get

#9y^2+25sqrt5+5=0#

One can also try for #cottheta=-1/2#, for which we get #sintheta=-2/sqrt5# and #costheta=1/sqrt5#

#x'=xcostheta+ysintheta# and #y'=-xsintheta-costheta#

i.e #x'=x/sqrt5-(2y)/sqrt5# and #y'=(2x)/sqrt5-y/sqrt5#

and simplifying #9x^2+25sqrt5=5=0#

Note - Please observe that above equation #x^2-4xy+4y^2+5sqrt5+1=0#

#hArr(x-2y)^2++5sqrt5+1=0# and as LHS for #x inRR# and #yinRR# is always positive, does not have real solution and as such cannot be represented on Cartesian Plane.

Dec 3, 2016

See below.

Explanation:

This possible conic can be written as

#C->(x,y)((1,-2),(-2,4))((x),(y))+5sqrt(5)+1=0#

Making a coordinate change such that

#((X),(Y))=R((x),(y))=((costheta,-sintheta),(sintheta,costheta))((x),(y))# we get

#C_R->(X,Y)R.M.R^T((X),(Y))+5sqrt(5)+1=0# or

#C_R->(X,Y)M_R((X),(Y))+5sqrt(5)+1=0# with

#M_R=R.M.R^T# resulting in

#C_R->(5/2-3/2cos(2theta)+2sin(2theta))X^2-(4cos(2theta)+3sin(2theta))XY+(5/2+3/2cos(2theta)-2sin(2theta))Y^2+5sqrt(5)+1=0#

Choosing #theta# such that the coefficient of #XY# is null or

#4cos(2theta)+3sin(2theta)=0# we get

#theta=-1/2 arctan(4,3)#

Now substituting this value into the rotation we have

#C_R->5Y^2+5+sqrt(5)+1=0#

This final result shows that #C_R# is not a real conic because there is not real solution satisfying it.

Dec 3, 2016

Please check the original equation. I think there is something wrong.

Explanation:

Here is a reference on Rotation of Axes

The general form of a conic is:

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

For the given equation:

#A = 1#
#B = -4#
#C = 4#
#D = 0#
#E = 0#
#F = 5sqrt(5) + 1#

The angle of rotation is:

#theta = 1/2tan^-1(B/(C-A))#

#theta = 1/2tan^-1(-4/(4-1))#

#theta = 1/2tan^-1(-4/3)#

#theta ~~ -26.565^@#

Using equations from the reference:

#A' = (A + C)/2 + [(A - C)/2] cos(2θ) - B/2 sin(2θ)#
#A' = (1 + 4)/2 + [(1 - 4)/2] cos(-53.13) - -4/2sin(-53.13)#
#A' = 5/2 -3/2cos(-53.13) + 2sin(-53.13)#
#A' = 0#
#B' = 0#
#C' = (A + C)/2 + [(C - A)/2] cos 2θ + B/2 sin 2θ#
#C' = 3.2#
#D' = 0#
#E' = 0#
#F' = F = 5sqrt(5) + 1#

I think there is something wrong; #A' = 0# should not be.
When I try to graph either the original equation or the rotated equation, using Desmos.com, I get nothing.