Answer
(a) Parabola.
(b) $\phi=53.13^{\circ}$, $X=\pm1$ (two lines).
(c) See graph.
Work Step by Step
(a) Write the equation as $9x^2+24xy+16y^2-25=0$, we can identify $A=9, B=24, C=16$. The discriminant is $B^2-4AC=24^2-36\times16=0$, thus the graph of the equation will be a parabola.
(b) Step 1. To eliminate the xy-term, we need to rotate the axes with an angle $\phi$ given by $cot2\phi=\frac{A-C}{B}=\frac{9-16}{24}=-\frac{7}{24}$ which gives $2\phi=106.26^{\circ}$ and $\phi=53.13^{\circ}$
Step 2. With $tan2\phi=-\frac{24}{7}$, we get $cos2\phi=-\frac{7}{25}$ and $sin\phi=\sqrt {\frac{1-cos2\phi}{2}}=\frac{4}{5}$ and $cos\phi=\sqrt {\frac{1+cos2\phi}{2}}=\frac{3}{5}$
Step 3. Use the conversion formulas $x=Xcos\phi-Ysin\phi=\frac{1}{5}(3X-4Y)$ and $y=Xsin\phi+Ycos\phi=\frac{1}{5}(4X+3Y)$
Step 4. Plug the above relation into the original equation to get $9(\frac{1}{5}(3X-4Y))^2+24(\frac{1}{5}(3X-4Y))(\frac{1}{5}(4X+3Y))+16(\frac{1}{5}(4X+3Y))^2-25=0$,
Step 5. Multiply 25 to both sides and simplify the equation to get $625X^2-625=0$ or $X^2=1$ which gives $X=\pm1$ or two lines.
(c) See graph.
