Identify the conic section given by y2+2y=4x2+3 Find its vertexvertices and focusfoci

Answered: "Identify the conic section given by y2+2y=4x2+3 Find..."

nicekikah

2021-02-10

Identify the conic section given by

y^{2} + 2 y = 4 x^{2} + 3

Find its

\frac{vertex}{vertices} and \frac{focus}{foci}

unett

Skilled2021-02-11Added 119 answers

Step 1
We rewrite the equation as:

y^{2} + 2 y = 4 x^{2} + 3

y^{2} + 2 y - 4 x^{2} = 3

y^{2} + 2 y + 1 - 4 x^{2} = 3 + 1

{(y + 1)}^{2} - 4 x^{2} = 4

\frac{{(y + 1)}^{2}}{4} - \frac{4 x^{2}}{4} = 1

\frac{{(y + 1)}^{2}}{4} - \frac{x^{2}}{1} = 1

This is an equation of a hyperbola.
Step 2
Then we compare the equation with standard form.

\frac{{(y + 1)}^{2}}{4} - \frac{x^{2}}{1} = 1

\frac{{(y - k)}^{2}}{b^{2}} - \frac{{(x - h)}^{2}}{a^{2}} = 1

h = 0, k = - 1,

a^{2} = 1, b^{2} = 4

a - 1, b = 2

vertex = (h, k \pm b) = (0, - 1 \pm 2) = (0, 1), (0, - 3)

Foci = (h, k \pm \sqrt{a^{2} + b^{2}} = (0, - 1 \pm \sqrt{1 + 4} = (0, - 1 + \sqrt{5}), (0, - 1 - \sqrt{5})

Answer:

vertex = (0, 1), (0, 3)

Foci = (0, - 1 + \sqrt{5}), (0, - 1 - \sqrt{5})

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