Question

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`?

Answer 1

Convert the equation into 1 of 2 standard forms:

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

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

Then the desired information can be obtained by observation.

Explanation:

Given: `5x^2-4y^2-40x-16y-36=0`

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

`5x^2-4y^2-40x-16y+5h^2-4k^2=36+5h^2-4k^2`

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

`5x^2-40x+5h^2-4y^2-16y-4k^2=36+5h^2-4k^2`

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

`5(x^2-8x+h^2)-4(y^2+4y+k^2)=36+5h^2-4k^2" [3]"`

Set the x term in the right side of the pattern `(x-h)^2 = x^2-2hx + h^2` equal to the x term in equation [3]:

`-2hx = -8x`

Solve for h:

`h = 4`

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

`5(x - 4)^2-4(y^2+4y+k^2)=36+5(4)^2-4k^2" [4]"`

Set the y term in the right side of the pattern `(y-k)^2 = y^2-2ky + k^2` equal to the y term in equation [6]:

`-2ky = 4y`

`k = -2`

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

`5(x - 4)^2-4(y- (-2))^2=36+5(4)^2-4(-2)^2`

Combine the terms on the right:

`5(x - 4)^2-4(y- (-2))^2=100`

Divide both sides by 100:

`(x - 4)^2/20-(y- (-2))^2/25=1`

Write the denominators as squares:

`(x - 4)^2/(2sqrt5)^2-(y- (-2))^2/5^2=1" [5]"`

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

The center is: `(h,k)`

From equation [5] we see that `h = 4 and k = -2` which gives us the point `(4,-2)`

The vertices are:

`(h-a,k) and (h+a,k)`

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

`(4-2sqrt5,-2) and (4+2sqrt5,-2)`

respectively

The foci are:

`(h-sqrt(a^2+b^2),k) and (h-sqrt(a^2+b^2),k)`

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

`(4-3sqrt5,-2) and (4+3sqrt5,-2)` respectively

The equations of the the asymptotes are:

`y = -b/a(x-h)+k` and `y = b/a(x-h)+k`

Respectively:

`y = -sqrt5/2(x-4)-2` and `y = sqrt5/2(x-4)-2`