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

Apr 28, 2016

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

$\left(3 x + 4 y + a\right) \left(3 x - 4 y + b\right) = c$

Comparing coefficients in the expansion with the given coefficients,
$a + b = 6 , b - a = 40 \mathmr{and} a b - c = - 247$.
$S o , a = - 17 , b = 23 \mathmr{and} c = - 144$

Now, the standard form of the given equation is

$\left(3 x + 4 y - 17\right) \left(3 x - 4 y + 23\right) + 144 = 0$.

The asymptotes of this hyperbola are given by

$\left(3 x + 4 y - 17\right) = 0 \mathmr{and} \left(3 x - 4 y + 23\right) = 0$

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 $A = 3 \mathmr{and} B = 4. {B}^{2} = {A}^{2} \left({e}^{2} - 1\right) . S o , 16 = 9 \left({e}^{2} - 1\right) . e = \frac{5}{3}$.

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

The two vertices are $V \left(- 1 , 8\right) \mathmr{and} V ' \left(- 1 , 2\right)$

Foci are on the major axis and are distant Ae = 5, from the center C(-1, 5).
So, the two foci are $S \left(- 1 , 5 + A e\right) = \left(- 1 , 10\right) \mathmr{and} S ' \left(- 1 , 5 - A e\right) = \left(- 1 , 0\right)$

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