Bruno Schneider

2023-03-07

How to write the equation of the hyperbola given Foci: (0,-5),(0, 5) and vertices (0, -3), (0,3)?

Julissa Heath

Beginner2023-03-08Added 3 answers

There are two different kinds of hyperbolas: one where the line connecting its vertices and foci is horizontal, and the other where the line is vertical. This hyperbola is the type where a line drawn through its vertices and foci is vertical. We know this by observing that it is the y coordinate that changes when we move from a focus point to a vertex.

The general equation for this type of hyperbola is:

$\frac{(y-k)\xb2}{a\xb2}-\frac{(x-h)\xb2}{b\xb2}=1$

Observe that the x coordinate of the foci and the vertices is 0; this tells us that $h=0$

$\frac{(y-k)\xb2}{a\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

The value of k is the sum of y coordinate of the vertices divided by two:$k=\frac{-3+3}{2}=0$

$\frac{(y-0)\xb2}{a\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

Please notice that when $x=0,y=\pm 3$; this allows us to find the value of ""a"":

$\frac{(3-0)\xb2}{a\xb2}-\frac{(0-0)\xb2}{b\xb2}=1$

$\frac{(3-0)\xb2}{a\xb2}=1$

$(3-0)\xb2=(a\xb2)$

$a=3$

$\frac{(y-0)\xb2}{3\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

Let c = the distance between the the center point and focus = 5. The equation for the square of this distance helps us to find the value of b:

$c\xb2=a\xb2+b\xb2$

$5\xb2=3\xb2+b\xb2$

$b\xb2=25-9$

$b\xb2=16$

$b=4$

$\frac{(y-0)\xb2}{3\xb2}-\frac{(x-0)\xb2}{4\xb2}=1$

Simplifying a bit:

$\frac{y\xb2}{3\xb2}-\frac{x\xb2}{4\xb2}=1$

The general equation for this type of hyperbola is:

$\frac{(y-k)\xb2}{a\xb2}-\frac{(x-h)\xb2}{b\xb2}=1$

Observe that the x coordinate of the foci and the vertices is 0; this tells us that $h=0$

$\frac{(y-k)\xb2}{a\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

The value of k is the sum of y coordinate of the vertices divided by two:$k=\frac{-3+3}{2}=0$

$\frac{(y-0)\xb2}{a\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

Please notice that when $x=0,y=\pm 3$; this allows us to find the value of ""a"":

$\frac{(3-0)\xb2}{a\xb2}-\frac{(0-0)\xb2}{b\xb2}=1$

$\frac{(3-0)\xb2}{a\xb2}=1$

$(3-0)\xb2=(a\xb2)$

$a=3$

$\frac{(y-0)\xb2}{3\xb2}-\frac{(x-0)\xb2}{b\xb2}=1$

Let c = the distance between the the center point and focus = 5. The equation for the square of this distance helps us to find the value of b:

$c\xb2=a\xb2+b\xb2$

$5\xb2=3\xb2+b\xb2$

$b\xb2=25-9$

$b\xb2=16$

$b=4$

$\frac{(y-0)\xb2}{3\xb2}-\frac{(x-0)\xb2}{4\xb2}=1$

Simplifying a bit:

$\frac{y\xb2}{3\xb2}-\frac{x\xb2}{4\xb2}=1$