# How do you find the center and vertex for #9x^2-16y^2+18x+160y-247=0#?

##### 1 Answer

This represents a hyperbola. The center is at (-1, 5). The vertcies are #(-1, 9) and (-1, 10)

#### Explanation:

This is the most general method for any second degree equation.

There is no xy-term and the product of the coefficients of #x^2 and

y^2 = -144<0#. So this equation represents a hyperbola..

The equation has the form

Comparing coefficients in the expansion with the given coefficients,

Now, the standard form of the given equation is

The asymptotes of this hyperbola are given by

So, the center of the hyperbola is the point of intersection C(-1, 5)..

Shifting the origin to the center using X = x+1 and Y = y-5, the equation becomes

The semi-axes are

The major axis is Y-axis (x=-1) and the vertices are X= 0, Y=+-A=+-3.

Referred to given axes,

The two vertices are

Foci are on the major axis and are distant Ae = 5, from the center C(-1, 5).

So, the two foci are

Just to give the most general method, I have avoided looking for other easier methods, for particular cases..

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