Question

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

Answer 1

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

`(3x+4y+a)(3x-4y+b)=c`

Comparing coefficients in the expansion with the given coefficients,
`a+b=6, b-a=40 and ab-c=-247`.
`So, a=-17, b=23 and c=-144`

Now, the standard form of the given equation is

`(3x+4y-17)(3x-4y+23)+144=0`.

The asymptotes of this hyperbola are given by

`(3x+4y-17)=0 and (3x-4y+23) =0`

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 `Y^2/3^2-X^2/4^2=1`

The semi-axes are `A = 3 and B = 4. B^2=A^2(e^2-1). So, 16=9(e^2-1). e = 5/3`.

The major axis is Y-axis (x=-1) and the vertices are X= 0, Y=+-A=+-3.
Referred to given axes,`x=-1 and y=2, 8`.

The two vertices are `V(-1, 8) and V'(-1, 2)`

Foci are on the major axis and are distant Ae = 5, from the center C(-1, 5).
So, the two foci are `S(-1, 5+Ae)=(-1, 10) and S'(-1, 5-Ae)=(-1, 0)`

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