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)

Analyzing functions
ANSWERED
asked 2021-01-04
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) \(\displaystyle{\left({x}–{6}\right)}^{{2}}={4}{\left({y}–{2}\right)}\)

Answers (1)

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 \(\displaystyle={k}–{p}={2}–{1}={1}\)
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