Question

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!?

Answer 1

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,.

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.

Answer 2

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.

Answer 3

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.