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

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:
$\frac{\left(y-k\right)²}{a²}-\frac{\left(x-h\right)²}{b²}=1$
Observe that the x coordinate of the foci and the vertices is 0; this tells us that $h=0$
$\frac{\left(y-k\right)²}{a²}-\frac{\left(x-0\right)²}{b²}=1$
The value of k is the sum of y coordinate of the vertices divided by two:$k=\frac{-3+3}{2}=0$
$\frac{\left(y-0\right)²}{a²}-\frac{\left(x-0\right)²}{b²}=1$
Please notice that when $x=0,y=±3$; this allows us to find the value of ""a"":
$\frac{\left(3-0\right)²}{a²}-\frac{\left(0-0\right)²}{b²}=1$
$\frac{\left(3-0\right)²}{a²}=1$
$\left(3-0\right)²=\left(a²\right)$
$a=3$
$\frac{\left(y-0\right)²}{3²}-\frac{\left(x-0\right)²}{b²}=1$
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:
$c²=a²+b²$
$5²=3²+b²$
$b²=25-9$
$b²=16$
$b=4$
$\frac{\left(y-0\right)²}{3²}-\frac{\left(x-0\right)²}{4²}=1$
Simplifying a bit:
$\frac{y²}{3²}-\frac{x²}{4²}=1$

Do you have a similar question?