# How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola 5x^2-4y^2-40x-16y-36=0?

May 20, 2017

Convert the equation into 1 of 2 standard forms:

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [2]}$

Then the desired information can be obtained by observation.

#### Explanation:

Given: $5 {x}^{2} - 4 {y}^{2} - 40 x - 16 y - 36 = 0$

Add $36 + 5 {h}^{2} - 4 {k}^{2}$ to both sides of the equation:

$5 {x}^{2} - 4 {y}^{2} - 40 x - 16 y + 5 {h}^{2} - 4 {k}^{2} = 36 + 5 {h}^{2} - 4 {k}^{2}$

Move all of the x related terms (on the left) so that they are the first 3 terms:

$5 {x}^{2} - 40 x + 5 {h}^{2} - 4 {y}^{2} - 16 y - 4 {k}^{2} = 36 + 5 {h}^{2} - 4 {k}^{2}$

Remove a factor of 5 from the first 3 terms and a factor of -4 from the remaining terms:

$5 \left({x}^{2} - 8 x + {h}^{2}\right) - 4 \left({y}^{2} + 4 y + {k}^{2}\right) = 36 + 5 {h}^{2} - 4 {k}^{2} \text{ [3]}$

Set the x term in the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ equal to the x term in equation [3]:

$- 2 h x = - 8 x$

Solve for h:

$h = 4$

Substitute ${\left(x - 4\right)}^{2}$ for $\left({x}^{2} - 8 x + {h}^{2}\right)$ on the left side of equation [3] and 4 for h on the right side:

$5 {\left(x - 4\right)}^{2} - 4 \left({y}^{2} + 4 y + {k}^{2}\right) = 36 + 5 {\left(4\right)}^{2} - 4 {k}^{2} \text{ [4]}$

Set the y term in the right side of the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ equal to the y term in equation [6]:

$- 2 k y = 4 y$

$k = - 2$

Substitute ${\left(y - \left(- 2\right)\right)}^{2}$ for $\left({y}^{2} + 4 y + {k}^{2}\right)$ on the left side of equation [4] and -2 for k on the right side:

$5 {\left(x - 4\right)}^{2} - 4 {\left(y - \left(- 2\right)\right)}^{2} = 36 + 5 {\left(4\right)}^{2} - 4 {\left(- 2\right)}^{2}$

Combine the terms on the right:

$5 {\left(x - 4\right)}^{2} - 4 {\left(y - \left(- 2\right)\right)}^{2} = 100$

Divide both sides by 100:

${\left(x - 4\right)}^{2} / 20 - {\left(y - \left(- 2\right)\right)}^{2} / 25 = 1$

Write the denominators as squares:

${\left(x - 4\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} - {\left(y - \left(- 2\right)\right)}^{2} / {5}^{2} = 1 \text{ [5]}$

This matches equation [1]; the horizontal transverse axis type.

The center is: $\left(h , k\right)$

From equation [5] we see that $h = 4 \mathmr{and} k = - 2$ which gives us the point $\left(4 , - 2\right)$

The vertices are:

$\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$

substitute in the values from equation [5]: gives us

$\left(4 - 2 \sqrt{5} , - 2\right) \mathmr{and} \left(4 + 2 \sqrt{5} , - 2\right)$

respectively

The foci are:

$\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right) \mathmr{and} \left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$

Substitute in the values from equation [5] gives us:

$\left(4 - 3 \sqrt{5} , - 2\right) \mathmr{and} \left(4 + 3 \sqrt{5} , - 2\right)$ respectively

The equations of the the asymptotes are:

$y = - \frac{b}{a} \left(x - h\right) + k$ and $y = \frac{b}{a} \left(x - h\right) + k$

Respectively:

$y = - \frac{\sqrt{5}}{2} \left(x - 4\right) - 2$ and $y = \frac{\sqrt{5}}{2} \left(x - 4\right) - 2$