Answer
a. $3x'^2+y'^2=2$
b. $\frac{3x'^2}{2}+\frac{y'^2}{2}=1$
c. See graph.
Work Step by Step
a. From the given equation, we have $A=C=1, B=1$. Thus $cot\theta=\frac{A-C}{B}=0$, $2\theta=90^\circ$ and $\theta=45^\circ$.
Use the transformation formulas
$x=x'cos\theta-y'sin\theta=x'cos45^\circ-y'sin45^\circ=\frac{\sqrt 2}{2}(x'-y')$
$y=x'sin\theta+y'cos\theta=x'sin45^\circ+y'cos45^\circ=\frac{\sqrt 2}{2}(x'+y')$
The original equation becomes
$x^2+xy+y^2-1=(\frac{\sqrt 2}{2}(x'-y'))^2+(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+(\frac{\sqrt 2}{2}(x'+y'))^2-1=\frac{1}{2}(x'^2-2x'y'+y'^2)+\frac{1}{2}(x'^2-y'^2)+\frac{1}{2}(x'^2+2x'y'+y'^2)-1=\frac{3}{2}x'^2+\frac{1}{2}y'^2-1=0$
or
$3x'^2+y'^2=2$
b. We can express the equation involving x′ and y′ in the standard form of a conic section as $\frac{3x'^2}{2}+\frac{y'^2}{2}=1$
c. See graph.
