Answer
(a) $(X-1)^2=\frac{1}{4}(Y+4)$
(b) XY-coordinates, vertex $V(1, -4)$. focus $F(1, -\frac{63}{16})$
xy-coordinates: vertex $V(\frac{5\sqrt 2}{2}, -\frac{3\sqrt 2}{2})$, focus $F(\frac{79\sqrt 2}{32}, -\frac{63\sqrt 2}{32})$
(c) $Y=-\frac{65}{16}$
$y=x-\frac{65\sqrt 2}{16}$
Work Step by Step
(a) Rewrite the original equation as $2\sqrt 2x^2+4\sqrt 2xy+2\sqrt 2y^2-7x-9y=0$ which gives $A=2\sqrt 2, B=4\sqrt 2, C=2\sqrt 2$. To remove the xy-term, we need to rotate the axes with an angle $\phi$ which is given by $\cot2\phi=\frac{A-C}{B}=0$, or $2\phi=\frac{\pi}{2}, \phi=\frac{\pi}{4}$. Use the rotation formula:
$x=Xcos\phi-Ysin\phi=\frac{\sqrt 2}{2}(X-Y)$, $y=Xsin\phi+Ycos\phi=\frac{\sqrt 2}{2}(X+Y)$.
Replace the variable in the original equation to get:
$2\sqrt 2(\frac{\sqrt 2}{2}(X-Y))^2+4\sqrt 2(\frac{\sqrt 2}{2}(X-Y))(\frac{\sqrt 2}{2}(X+Y))+2\sqrt 2(\frac{\sqrt 2}{2}(X+Y))^2-7(\frac{\sqrt 2}{2}(X-Y))-9(\frac{\sqrt 2}{2}(X+Y))=0$
Multiply 2 on both sides and combine the like terms to get $8\sqrt 2X^2-16\sqrt 2X-2\sqrt 2Y=0$ which gives $(X-1)^2=\frac{1}{4}(Y+4)$ that is a parabola.
(b) For the XY-coordinates, the vertex is at $V(1, -4)$. With $4p=\frac{1}{4}$, we have $p=\frac{1}{16}$ and the focus is at $F(1, \frac{1}{16}-4)$ or $F(1, -\frac{63}{16})$
To find corresponding values in the xy-coordinates, use the conversion formula:
$x=Xcos\phi-Ysin\phi=\frac{\sqrt 2}{2}(X-Y)$, $y=Xsin\phi+Ycos\phi=\frac{\sqrt 2}{2}(X+Y)$.
The vertex is at $V(\frac{\sqrt 2}{2}(1+4), \frac{\sqrt 2}{2}(1-4))$ or $V(\frac{5\sqrt 2}{2}, -\frac{3\sqrt 2}{2})$
and the focus is at $F(\frac{\sqrt 2}{2}(1+\frac{63}{16}), \frac{\sqrt 2}{2}(1-\frac{63}{16}))$ or $F(\frac{79\sqrt 2}{32}, -\frac{63\sqrt 2}{32})$
(c) The equation of the directrix in XY-coordinates is given by $Y=-4-\frac{1}{16}$ or $Y=-\frac{65}{16}$
To get the corresponding equation in the xy-coordinates, use the conversion:
$X=x\cdot cos\phi+y\cdot sin\phi=\frac{\sqrt 2}{2}(x+y)$ and $Y=-x\cdot sin\phi+y\cdot cos\phi=\frac{\sqrt 2}{2}(-x+y)$
We can get the directrix as $\frac{\sqrt 2}{2}(-x+y)=-\frac{65}{16}$ or $y=x-\frac{65\sqrt 2}{16}$
