Question

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).

Answer 1

The standard form of the hyperbola is \[\frac{ x^2 }{ a^2 }-\frac{ y^2 }{ b }=1\] where a>b and the foci is (k,0)\[k^2=a^2+b^2\]

Answer 2

Vertices would be (a,0) and (-a,0)

Answer 3

So it would be \[\frac{ 4^2 }{ 121}-\frac{ x^2 }{ 4 }=1\]

Answer 4

correction; it should be x^2/a^2-y^2/b^2=1

Answer 5

That is if the hyperbola is drawn like this.

Answer 6

Notice that the foci for the question is on the y-axis, not x-axis, right?

Answer 7

Another form of the hyperbola is \[\frac{ y^2 }{ b^2 }-\frac{ x^2 }{ a^2 }=1\]
foci are (0,k)\[a^2+b^2=k^2\]
The verticies are (0,b) and (0,-b)

(b>a)

Answer 8

We have to use the second form since the foci are on the y-axis

Answer 9

a^2 is 2^2 and b^2 is 11^2?

Answer 10

First, we can find b since the vertices are given.

\[(0,\pm b)=(0,\pm2)\rightarrow b=2\]

Answer 11

Then, we can find a\[(0, \pm k)=(0,\pm11)\rightarrow k=11\]

Answer 12

Because a^2+b^2=k^2...
\[11^2=a^2+b^2=a^2+2^2\rightarrow a^2=117\]

Answer 13

The standard form of the equation for this type of hyperbola is:
(y^2 / a^2) - (x^2 / b^2) =1
Your coordinates of the vertices are (0, +/-2) and (0 +/-11) making a=2 and c=11.
To find the equation of the vertical hyperbola, we'll first need to use the 'a' and 'c' values to find 'b' using the formula b^2=C^2 - a^2.
b^2 = 117
(y^2/4) - (x^2/117) = 1

Answer 14

117y^2 -4x^2 = 468

Answer 15

So it's final equation :D hope you get it