Question

What is the equation of a hyperbola with foci at (2, 8) and (2, 2) and vertices at (2, 7) and (2, 3)?

Answer 1

Answer:

$\frac{(x-2)^2}{4}+\frac{(y-5)^2}{5}=1..$

Step-by-step explanation:

Major axis is the line joining the vertices A(2, 3) and A'(2, 7), x = 2..

The axes of the hyperbola are parallel to the axes of coordinates.

Distance between vertices AA' = major-axis length = 2 a = 4. a = 2..

The center C is the midpoint of AA'. So, C is (2, 5)

The distance between foci S(2, 2) and S'(2, 8), SS' = 2 a (eccentricity) = 2 a e = 6.

So, $4e=6. e=\frac{3}{2}$

The semi-transverse axis is $b=a\sqrt{e^2-1}=\sqrt{5}$

The equation required is the equation of the hyperbola with center at C(2, 5), axes parallel to axes of coordinates and semi-axes $a=2$ and $b=\sqrt{5}$ is....

$\frac{(x-2)^2}{4}+\frac{(y-5)^2}{5}=1..$