Find the center, vertices, the foci and the asymptoted of the hyperbola.?

Question

x^2 - y^2 - 4x - 2y + 2 = 0.

The center is ___

The vertices are ___

The foci are ___

The asymptoted are y = ___ and y = ___

Answer 1

First, put it in the general form, which is:

(x - h)²/a² - (y - k)²/b² = 1

x² - 4x - y² - 2y = -2

x² - 4x + 4 - (y² + 2y + 1) = -2 + 4 - 1

(x - 2)² - (y + 1)² = 1

That's the general form. It's a parabola centered at (2, 1) opening "east-west". a = b = 1.

So....

The center is (2, 1).

To find the vertices, you add and subtract the "semimajor axis" (aka a) to the center, in the direction that the hyperbola opens (x for east-west, y for north-south).

So the vertices are (2 + 1, 1) or (3, 1), and (2 - 1, 1) or (1, 1).

To find the foci, first you have to compute another value (whose name I don't know), called c:

c² = a² + b² = 1 + 1 = 2, so c = √2

And then to get the foci, you add and subtract c to the center in the same way as the vertices:

The foci are (2 + √2, 1) and (2 - √2, 1).

And the asymptotes for an east-west hyperbola are y = ± b/a, and for a north-south hyperbola are y = ±a/b.

This is an east-west hyperbola, so the asymptotes are:

y = b/a x = 1/1x = x and
y = -b/a x = -1/1x = -x