Answer
(a) ellipse.
(b) $\phi\approx36.9^{\circ}$.
$X^2+\frac{Y^2}{9}=1$
(c) See graph.
Work Step by Step
(a) Rewrite the original equation as $153x^2+192xy+97y^2+0x+0y-225=0$ and we have $A=153, B=192, C=97, D=E=0,F=-225$. The discriminant is $B^2-4AC=192^2-4\times153\times97=-22500\lt0$, thus the graph of the equation is an ellipse.
(b) To eliminate the xy-term, we need a rotation of axes with an angle $\phi$ where $cot2\phi=\frac{A-C}{B}=\frac{153-97}{192}=\frac{7}{24}$ which gives $2\phi\approx73.74^{\circ}$ and $\phi\approx36.9^{\circ}$.
With $\tan2\phi=\frac{24}{7}$, we have $cos2\phi=\frac{7}{25}$. With $\phi$ in Quadrant I, we have $sin\phi=\sqrt {\frac{1-cos2\phi}{2}}=\frac{3}{5}$ and $cos\phi=\frac{4}{5}$
The transformation formula gives:
$x=X\cdot \frac{4}{5} - Y\cdot\frac{3}{5}=\frac{1}{5}(4X-3Y)$,
$y=X\cdot \frac{3}{5}+Y\cdot \frac{4}{5}=\frac{1}{5}(3X+4Y)$
Use them in the original equation to get $153(\frac{1}{5}(4X-3Y))^2+192(\frac{1}{5}(4X-3Y))(\frac{1}{5}(3X+4Y))+97(\frac{1}{5}(3X+4Y))^2-225=0$
Simplify this equation by multiplying 25 and combine like terms to get $5625X^2+625Y^2=25\times225$ or
$X^2+\frac{Y^2}{9}=1$
(c) See graph.
