The hyperbolass

Question

Find an equation for the hyperbola described. Foci at (9,6) and (9,10); Vertex (9,9)

Answer 1

hyperbola with vertex (9,9) foci (9,6) and (9,10)

has axes of symmetry along the lines x=9 and y=8

center midway between the foci at (9, (10+6)/2) = (9, 8)

it has a 2nd vertex at (9,7)

it's a vertical hyperbola

(y-k)/a^2 - (x-h)^2/b^2 = 1 with (h,k) the center = (9,8)

(y-7.5)^2/a^2 - (x-9)^2/b^2 = 1 where a=distance from center to vertex = 1

c= distance from center to focus = 2

b^2 = c^2 -a^2 = 4-1 = 3

b = sqr3

(y-8)^2/(1)^2 - (x-9)^2/(sqr3)^2 = 1, in standard form

or (y-8)^2 - (x-9)^2/3 = 1

or removing fractions

3(y^2 -16y +64) -x^2 +18x -81 = -3

x^2 -18x -3y^2 +48y = -3+192-81=114

x^2 -18x -3y^2 +48y = 108

it may help to try sketching the graph. It has two branches, one above the other

symmetrical about the line y=y coordinate of the center: y=8

with 2 asymptotes: y =+/-a/b= +/-1/sqr3

when the x and y variables both have a squared term the graph is either a hyperbola or an ellipse

if the two squared terms are opposite signs, it's a hyperbola. if they're the same sign it's an ellipse

Answer 2

We can observe that the foci and the one vertex are located on the same vertical line x=9, so we need to use the formula for a vertical hyperbola which is (y-k)²/a² - (x-h)²/b² = 1. You can remember this formula because 2a is longer than 2b in terms of the vertical y-axis. We want to use the relationship a²+b²=c² to help us.

To calculate a, we need to find the distance from the center to the vertex. The center is easily (h,k)=(9,8) because it is equidistant from both foci (which is actually c=2). So, the distance from the center (9,8) to the vertex (9,9) is a=1.

Therefore, we can determine b² using our values of a and c:

a²+b²=c²

1²+b²=2²

1+b²=4

b² = 3

Hence, our hyperbola that has foci at (9,6) and (9,10) with vertex (9,9) is (y-8)²/1 - (x-9)²/3 = 1

Hope this helped!

Answer 3

2c = 10 - 6 = 4; c = 2; a = 10 - 9 = 1; Center at (9, 8); b² = 4 - 1 = 3.

Equation of hyperbola (y - 8)²/1 - (x - 9)²/3 = 1