Question

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

Answer 1

Use the standard form for a hyperbola of this type:

`(y-k)^2/a^2-(x-h)^2/b^2 = 1`

The vertices are: `(h, k-a)` and `(h,k+a)`
And the foci are: `(h, k-sqrt(a^2+b^2))` and `(h, k+sqrt(a^2+b^2))`

Explanation:

Given: `4y^2 - 9x^2 - 40y - 54x - 17 = 0" [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:

`(y-k)^2/a^2-(x-h)^2/b^2 = 1" [2]"`

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

The vertices are: `(h, k-a)` and `(h,k+a)`
And the foci are: `(h, k-sqrt(a^2+b^2))` and `(h, k+sqrt(a^2+b^2))`

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

`4y^2 - 40y - 9x^2 - 54x = 17" [1.1]"`

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

`4y^2 - 40y+4k^2 - 9x^2 - 54x-9h^2 = 17+4k^2-9h^2" [1.2"`

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

`4(y^2 - 10y+k^2) - 9(x^2 + 6x+h^2) = 17+4k^2-9h^2" [1.3]"`

From the expansion `(y-k)^2 = y^2-2ky+k^2`, we can write the equation:

`-2ky = -10y`

`k = 5`

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

`4(y - 5)^2 - 9(x^2 + 6x+h^2) = 17+4(5)^2-9h^2" [1.4]"`

From the expansion `(x-k)^2 = x^2-2hx+h^2`, we can write the equation:

`-2hx = 6x`

`h = -3`

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

`4(y - 5)^2 - 9(x - (-3))^2 = 17+4(5)^2-9(-3)^2" [1.5]"`

Simplify the right side of equation [1.5]:

`4(y - 5)^2 - 9(x - (-3))^2 = 36" [1.5]"`

Divide both sides of the equation by 36:

`(y - 5)^2/9 - (x - (-3))^2/4 = 1" [1.6]"`

Write the denominators as squares:

`(y - 5)^2/3^2 - (x - (-3))^2/2^2 = 1" [1.7]"`

Using the pattern for the vertices, `(h, k-a)` and `(h,k+a)`

`(-3,5-3)` and `(-3, 5+3)`

The vertices are: `(-3,2)` and `(-3, 8)`

Using the pattern for the foci, `(h, k-sqrt(a^2+b^2))` and `(h, k+sqrt(a^2+b^2))`

`sqrt(a^2+b^2) = sqrt(9+4) = sqrt(13)`

The foci are: `(-3, 5-sqrt(13))` and `(-3, 5+sqrt(13))`