Write equation in standard form hyperbola given endpoints and foci.

Question

Endpoints (-5,-4) & (-5,6)

foci (-5,-5) & (-5,7)

Answer 1

(y-1)^2/25 -(x+5)^2/11 = 1

(y-k)^2/a^2 -(x-h)^2/b^2 = 1 is the standard equation for a hyperbola whose transverse axis is parallel to the y-axis, where (h,k) is the center, the midpoint between the endpoints (vertices) of the hyperbola. k=(6-4)/2= 1

(h,k) = (-5,1)

a=5

b=sqr11

b=sqr(c^2-a^2)

c=-6

the foci are (h,k+c) and (h,k-c) = (-5,-5) and (-5,7)

k+c = 1+c =-5

c =-5-1=-6

b^2 =36-25=11

b=sqr11

You could also try to get the equation by using the definition of a hyperbola as the locus of all points such that their difference in distance from the 2 foci are equal. In this case, take either vertex and calculate the distance to each focus, 1 and 11. Their difference = 10. The hyperbola is all (x,y) points such that their distance from one focus minus the distance to the other =10.

Answer 2

The center of the hyperbola is (-5, 1). The hyperbola is vertical, so our equation (so far) is

(y - 1)^2/a^2 - (x+ 5)^2/b^2 = 1. Remember that we have the relationship c^2 = a^2 + b^2.

The distance from the center to a vertex is 5, which is the value of a. So the vertices are at (-5, 6) and (-5, -4).

The distance from the center to either focus is 6, which is the value of c.

So c^2 = a^2 + b^2 is 6^2 = 5^2 + b^2. 11 = b^2

The equation is now: (y-1)^2/25 - (x+5)^2/11 = 1

If you need to write this out without the fractions:

multiply the equation by the common denominator 275.

The equation becomes 11y^2 - 22y - 25x^2 - 250x - 889 = 0.

Or -25x^2 + 11y^2 - 250x - 22y - 250x - 880 = 0