nicekikah

2021-02-10

Identify the conic section given by ${y}^{2}+2y=4{x}^{2}+3$

Find its$\frac{\text{vertex}}{\text{vertices}}\text{}\text{and}\text{}\frac{\text{focus}}{\text{foci}}$

Find its

unett

Skilled2021-02-11Added 119 answers

Step 1

We rewrite the equation as:

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

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

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

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

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

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

This is an equation of a hyperbola.

Step 2

Then we compare the equation with standard form.

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

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

$h=0,k=-1,$

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

$a-1,b=2$

$\text{vertex}\text{}{\displaystyle =(h,k\pm b)=(0,-1\pm 2)=(0,1),(0,-3)}$

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

Answer:

$\text{vertex}\text{}{\displaystyle =(0,1),(0,3)}$

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

We rewrite the equation as:

This is an equation of a hyperbola.

Step 2

Then we compare the equation with standard form.

Answer: