Find an equation for the conic section with the given properties. The hyperbola with vertices V1(−1, −3) and V2(7, −3) and foci F1(−5, −3) and F2(11, −3)

Answer 1

Note that the y coordinates of the foci and vertices are all the same, -3. Thus, distance may be calculated by subtracting x values, and the hyperbola will be of the form

(x-h)²/a2 + (y-k)2/b² = 1, where

(h, k) is the center

a is the distance from the vertex to the center

c is the distance from the focus to the center

b = √(c² - a²)

For the center, k, the y co-ordinate, equals -3

h, the x co-ordinate, = (7 + -1)/2 = 3; you may average either the vertices or foci

a = 7 - 3 = 4 - distance from the vertex to the center, as y co-ordinates are the same

c = 11 - 3 = 8 - distance from the foci to the center.

b = √(c2 - a²) = √(8² - 4²) = 4√(2² - 1²) = 4√3

Now, we can plug in and have the equation for the focus. Note that a² = 16 and b² = 48

Then, (x-3)²/16 - (y+3)²/48 = 1

A hyperbola may be defined as the locus of points such that the difference of their distance to the 2 foci is a constant; the distance equals the distance between the vertices.

An ellipse has the same definition except difference is replaced by sum.

This gives a geometric meaning to the foci.