# How do you graph & find the focus as well as the vertex of 4y^2 - 9x^2 - 40y - 54x - 17 = 0?

Jun 25, 2017

Use the standard form for a hyperbola of this type:

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

The vertices are: $\left(h , k - a\right)$ and $\left(h , k + a\right)$
And the foci are: $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right)$ and $\left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

#### Explanation:

Given: $4 {y}^{2} - 9 {x}^{2} - 40 y - 54 x - 17 = 0 \text{ [1]}$

Here is a graph of equation [1].

graph{4y^2 - 9x^2 - 40y - 54x - 17 = 0 [-20, 20, -6, 16]}

The standard form for a hyperbola of this type is:

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

We want to make equation [1] look like equation [2], because we know that:

The vertices are: $\left(h , k - a\right)$ and $\left(h , k + a\right)$
And the foci are: $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right)$ and $\left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

Group the x terms and y terms together and move the constant to the right:

$4 {y}^{2} - 40 y - 9 {x}^{2} - 54 x = 17 \text{ [1.1]}$

We want to complete the squares in equation [1.1], therefore, we shall add $4 {k}^{2} - 9 {h}^{2}$ to both sides of the equation:

$4 {y}^{2} - 40 y + 4 {k}^{2} - 9 {x}^{2} - 54 x - 9 {h}^{2} = 17 + 4 {k}^{2} - 9 {h}^{2} \text{ [1.2}$

Remove a common factor of 4 from the y terms and a common factor of -9 from the x terms:

$4 \left({y}^{2} - 10 y + {k}^{2}\right) - 9 \left({x}^{2} + 6 x + {h}^{2}\right) = 17 + 4 {k}^{2} - 9 {h}^{2} \text{ [1.3]}$

From the expansion ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, we can write the equation:

$- 2 k y = - 10 y$

$k = 5$

Into equation [1.3], substitute ${\left(y - 5\right)}^{2}$ on the left and 4(5)^2 on the right:

$4 {\left(y - 5\right)}^{2} - 9 \left({x}^{2} + 6 x + {h}^{2}\right) = 17 + 4 {\left(5\right)}^{2} - 9 {h}^{2} \text{ [1.4]}$

From the expansion ${\left(x - k\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, we can write the equation:

$- 2 h x = 6 x$

$h = - 3$

Into equation [1.4], substitute ${\left(x - \left(- 3\right)\right)}^{2}$ on the left and -9(-3)^2 on the right:

$4 {\left(y - 5\right)}^{2} - 9 {\left(x - \left(- 3\right)\right)}^{2} = 17 + 4 {\left(5\right)}^{2} - 9 {\left(- 3\right)}^{2} \text{ [1.5]}$

Simplify the right side of equation [1.5]:

$4 {\left(y - 5\right)}^{2} - 9 {\left(x - \left(- 3\right)\right)}^{2} = 36 \text{ [1.5]}$

Divide both sides of the equation by 36:

${\left(y - 5\right)}^{2} / 9 - {\left(x - \left(- 3\right)\right)}^{2} / 4 = 1 \text{ [1.6]}$

Write the denominators as squares:

${\left(y - 5\right)}^{2} / {3}^{2} - {\left(x - \left(- 3\right)\right)}^{2} / {2}^{2} = 1 \text{ [1.7]}$

Using the pattern for the vertices, $\left(h , k - a\right)$ and $\left(h , k + a\right)$

$\left(- 3 , 5 - 3\right)$ and $\left(- 3 , 5 + 3\right)$

The vertices are: $\left(- 3 , 2\right)$ and $\left(- 3 , 8\right)$

Using the pattern for the foci, $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right)$ and $\left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{9 + 4} = \sqrt{13}$

The foci are: $\left(- 3 , 5 - \sqrt{13}\right)$ and $\left(- 3 , 5 + \sqrt{13}\right)$