Find an equation in standard form for the hyperbola that satisfies the given conditions. Foci ​(−5​,5​) and ​(3​,5​), transverse axis endpoints ​(−4​,5​) and ​(2​,5​).

Question

Find an equation in standard form for the hyperbola that satisfies the given conditions. Foci ​(−5​,5​) and ​(3​,5​), transverse axis endpoints ​(−4​,5​) and ​(2​,5​).

Answer 1

(x+1)^2/9 - (y-5)^2/7 = 1

7(x+1)^2 - 9(y-5)^2 = 7(9) = 63

7x^2 +14x + 7 -9y^2 +90y -225 = 63

7x^2 +14x -9y^2 + 90y = 63+225-7 = 169

7x^2 +14x -9y^2 +90y = 169

the very 1st equation above is a standard form

last equation eliminates fractions, but it is the same hyperbola

all the above are the same hyperbola

Answer 2

It really helps to plot out the information on a graph in order to get a visual of where everything. If you plot the foci and end points you can see the traverse axis is in the horizontal direction. So the standard equation is in this form:

(x-h)²/a²-(y-k)²/b² = 1

The center (h,k) can be seen as the midpoint for both the foci and endpoints. The center is (-1,5)

You can determine the the distance from the center to the endpoint is 3, so a=3. You can also determine the the distance from the center to the foci is 4, so c=4. If we know a and c, we can calculate b.

c²=a²+b² Plugging in values

4²=3²+b²

16=9+b²

7=b²

Plugging everything in:

(x+1)²/9 - (y-5)²/7= 1