Consider the equation of a parabola $\mathrm{x}^{2}-4 \mathrm{x}-4 \mathrm{y}+8=0$. Find the focus, vertex, axis of symmetry, and the directrix.

Show that if the pair of numbers \((\mathrm{x}, \mathrm{y})\) satisfies \(\mathrm{y}=(1 / 4) \mathrm{x}^{2}\), then the distance FP from \(\mathrm{F}(0,1)\) to \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is equal to the distance \(\mathrm{PQ}\) from \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) to \(\mathrm{Q}(\mathrm{x},-1)\).

Find, both analytically and graphically, the points of intersection of the two curves whose equations are $$ 2 x+y-4=0 \text { and } y^{2}-4 x=0 $$

By definition, if an hyperbola has foci \(F_{1}(-c, 0) F_{2}(c, 0)\), and \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is a point on the hyperbola, then \(\left|\mathrm{PF}_{1}-\mathrm{PF}_{2}\right|=\mathrm{k}\), where \(\mathrm{k}\) is a constant such that $\mathrm{k}<\mathrm{F}_{1} \mathrm{~F}_{2}=2 \mathrm{c}$ Assuming that the above holds, and defining a such that $\mathrm{a}=\mathrm{K} / 2\(. and a constant \)\mathrm{b}$ such that \(\mathrm{b}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) prove that the equation of the hyperbola is $$ \left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1 $$

Find the equation of a parabola that has vertex at \((-1,2)\), axis of symmetry parallel to the \(\mathrm{x}\) -axis, and goes through the point \(\mathrm{P}_{1}(-3,-4)\).

Show that if the coordinates \((\mathrm{x}, \mathrm{y})\) of a point \(\mathrm{P}\) satisfy $$ \left(x^{2} / 9\right)-\left(y^{2} / 16\right)=1 $$ then \(\left|\mathrm{F}_{1} \mathrm{P}-\mathrm{F}_{2} \mathrm{P}\right|=6\), where \(\mathrm{F}_{1}(-5,0)\) and \(\mathrm{F}_{2}(5,0)\) are the foci.