Question

Identify the conic section given by \(\displaystyle{y}^{2}+{2}{y}={4}{x}^{2}+{3}\)
Find its \(\frac{\text{vertex}}{\text{vertices}}\ \text{and}\ \frac{\text{focus}}{\text{foci}}\)

Answer 1

Step 1
We rewrite the equation as:
\(\displaystyle{y}^{2}+{2}{y}={4}{x}^{2}+{3}\)
\(\displaystyle{y}^{2}+{2}{y}-{4}{x}^{2}={3}\)
\(\displaystyle{y}^{2}+{2}{y}+{1}-{4}{x}^{2}={3}+{1}\)
\(\displaystyle{\left({y}+{1}\right)}^{2}-{4}{x}^{2}={4}\)
\(\displaystyle\frac{{{\left({y}+{1}\right)}^{2}}}{{4}}-\frac{{{4}{x}^{2}}}{{4}}={1}\)
\(\displaystyle\frac{{{\left({y}+{1}\right)}^{2}}}{{4}}-\frac{{x}^{2}}{{1}}={1}\)
This is an equation of a hyperbola.
Step 2
Then we compare the equation with standard form.
\(\displaystyle\frac{{{\left({y}+{1}\right)}^{2}}}{{4}}-\frac{{x}^{2}}{{1}}={1}\)
\(\displaystyle\frac{{{\left({y}-{k}\right)}^{2}}}{{b}^{2}}-\frac{{{\left({x}-{h}\right)}^{2}}}{{a}^{2}}={1}\)
\(\displaystyle{h}={0},{k}=-{1},\)
\(\displaystyle{a}^{2}={1},{b}^{2}={4}\)
\(\displaystyle{a}-{1},{b}={2}\)
\(\text{vertex}\ \displaystyle={\left({h},{k}\pm{b}\right)}={\left({0},-{1}\pm{2}\right)}={\left({0},{1}\right)},{\left({0},-{3}\right)}\)
\(\text{Foci}\ \displaystyle={\left({h},{k}\pm\sqrt{{{a}^{2}+{b}^{2}}}={\left({0},-{1}\pm\sqrt{{{1}+{4}}}={\left({0},-{1}+\sqrt{{5}}\right)},{\left({0},-{1}-\sqrt{{5}}\right)}\right.}\right.}\)
Answer:
\(\text{vertex}\ \displaystyle={\left({0},{1}\right)},{\left({0},{3}\right)}\)
\(\text{Foci}\ \displaystyle={\left({0},-{1}+\sqrt{{5}}\right)},{\left({0},-{1}-\sqrt{{5}}\right)}\)