Answer
(a) parabola.
(b) $\phi\approx22.62^{\circ}$.
$169Y^2+120Y=119X$
(c) See graph.
Work Step by Step
(a) Rewrite the original equation as $25x^2-120xy+144y^2-156x-65y=0$ and we have $A=25, B=-120, C=144, D=-156,E=-65,F=0$. The discriminant is $B^2-4AC=120^2-4\times25\times144=0$, thus the graph of the equation is a parabola.
(b) To eliminate the xy-term, we need a rotation of axes with an angle $\phi$ where $cot2\phi=\frac{A-C}{B}=\frac{25-144}{-120}=\frac{119}{120}$ which gives $2\phi\approx45.24^{\circ}$ and $\phi\approx22.62^{\circ}$.
With $tan2\phi=\frac{120}{119}$, we have $cos2\phi=\frac{119}{169}$. Thus $sin\phi=\sqrt {\frac{1-cos2\phi}{2}}=\frac{5}{13}$ and $cos\phi=\sqrt {\frac{1+cos2\phi}{2}}=\frac{12}{13}$
The transformation formula gives:
$x=X\cdot \frac{12}{13} - Y\cdot \frac{5}{13}=\frac{1}{13}(12X-5Y)$,
$y=X\cdot \frac{5}{13}+Y\cdot \frac{12}{13}=\frac{1}{13}(5X+12Y)$
Use them in the original equation to get $25(\frac{1}{13}(12X-5Y))^2-120(\frac{1}{13}(12X-5Y))(\frac{1}{13}(5X+12Y))+144(\frac{1}{13}(5X+12Y))^2-156(\frac{1}{13}(12X-5Y))-65(\frac{1}{13}(5X+12Y))=0$
Simplify this equation by multiplying 169 and combine the llike terms to get $28561Y^2-20111X+20280Y=0$ or $169Y^2+120Y-119X=0$ which gives $(Y+\frac{60}{169})^2=\frac{119}{169}(X+\frac{3600}{20111})$
(c) See graph.
