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Question # Find the vertex, focus, and directrix for the parabolas: a) (y – 9)^2 = 8(x -2) b) y^2 – 4y = 4x – 2^2 c) (x – 6)^2 = 4(y – 2)

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ANSWERED Find the vertex, focus, and directrix for the parabolas:
a) $$\displaystyle{\left({y}–{9}\right)}^{{2}}={8}{\left({x}-{2}\right)}$$
b) $$\displaystyle{y}^{{2}}–{4}{y}={4}{x}–{2}^{{2}}$$
c) $$\displaystyle{\left({x}–{6}\right)}^{{2}}={4}{\left({y}–{2}\right)}$$ 2021-01-05
a) $$\displaystyle{\left({y}–{9}\right)}^{{2}}={8}{\left({x}-{2}\right)}$$
$$\displaystyle{\left({y}–{k}\right)}^{{2}}={4}{p}{\left({x}–{h}\right)}$$
$$\displaystyle{h}={2},{k}={9},$$
$$\displaystyle{4}{p}={8}$$
$$\displaystyle{p}=\frac{{8}}{{4}}={2}$$
vertex $$\displaystyle={\left({h},{k}\right)}={\left({2},{9}\right)}$$
focus $$\displaystyle={\left({4},{9}\right)}$$
directrix $$\displaystyle={x}={h}–{p}={2}–{2}={0}$$
b) $$\displaystyle{y}^{{2}}–{4}{y}={4}{x}–{2}^{{2}}$$ – transform, as $$\displaystyle{a}^{{2}}+{b}^{{2}}+{2}{a}{b}={\left({a}+{b}\right)}^{{2}}$$, so
$$\displaystyle{y}^{{2}}–{4}{y}+{4}={4}{x}+{4}+{4}$$
$$\displaystyle{\left({y}–{2}\right)}^{{2}}={4}{\left({x}+{2}\right)}$$
$$\displaystyle{h}=-{2},{k}={2}$$
$$\displaystyle{4}{a}={4}$$
$$\displaystyle{a}={1}$$
vertex $$\displaystyle={\left({h},{k}\right)}={\left(-{2},{2}\right)}$$
focus $$\displaystyle={\left(-{2},{3}\right)}$$
directx $$\displaystyle={h}–{p}=-{2}–{3}=-{5}$$
c) $$\displaystyle{\left({x}–{6}\right)}^{{2}}={4}{\left({y}–{2}\right)}$$
$$\displaystyle{\left({x}–{h}\right)}^{{2}}={4}{p}{\left({y}–{k}\right)}$$
$$\displaystyle{h}={6},{k}={2}$$
$$\displaystyle{4}{p}={4}$$
$$\displaystyle{p}={1}$$
vertex $$\displaystyle={\left({h},{k}\right)}={\left({6},{2}\right)}$$
focus $$\displaystyle={\left({h},{k}+{p}\right)}={\left({6},{3}\right)}$$
directrix $$\displaystyle={k}–{p}={2}–{1}={1}$$