How do you rotate the axes to transform the equation `x^2+2sqrt3xy-y^2=7` into a new equation with no xy term and then find the angle of rotation?

A conic equation of the type of `Ax^2+Bxy+Cy^2+Dx+Ey+F=0` is rotated by an angle `theta`, to form a new Cartesian plane with coordinates `(x',y')`, if `theta` is appropriately chosen, we can have a new equation without term `xy` i.e. of standard form.
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The relation between coordinates `(x,y)` and `(x'.y')` can be expressed as
`x=x'costheta-y'sintheta` and `y=x'sintheta+y'costheta`

or `x'=xcostheta+ysintheta` and `y=-xsintheta+ycostheta`

for this we need to have `theta` given by `cot2theta=(A-C)/B`

In the given case as equation is `x^2+2sqrt3xy-y^2=7`, we have `A=1`, `B=2sqrt3` and `C=-1` and hence `cot2theta=2/(2sqrt3)=1/sqrt3` i.e. `theta=pi/6`

Hence relation is give by `x=x'cos(pi/6)-y'sin(pi/6)` and `y=x'sin(pi/6)+y'cos(pi/6)` i.e.

`x=(x'sqrt3)/2-(y')/2` and `y=(x')/2+(y'sqrt3)/2`

and `x^2+2sqrt3xy-y^2=7` is

`((x'sqrt3)/2-(y')/2)^2+2sqrt3((x'sqrt3)/2-(y')/2)((x')/2+(y'sqrt3)/2)-((x')/2+(y'sqrt3)/2)^2=7`

which simplifies to `2(x'^2-y'^2)=7` or we can say

`2x^2-2y^2-7=0`

graph{(2x^2-2y^2-7)(x^2+2sqrt3xy-y^2-7)=0 [-10, 10, -5, 5]}