Question

How do you identify the vertices, foci, and direction of `-10y-y^2=-4x^2-72x-199`?

Answer 1

Write the equation in one of the two standard forms then use information to find the vertices and foci.

Explanation:

Given:

`-10y-y^2=-4x^2-72x-199`

Add `4x^2+72x-k^2+4h^2` to both sides of the equation:

`4x^2+72x+4h^2-y^2-10y-k^2=-199+4h^2-k^2`

Remove a common factor of 4 from the first 3 terms and a common factor of -1 from the next 3 terms:

`4(x^2+18x+h^2)-(y^2+10y+k^2)=-199+4h^2-k^2" [1]"`

Use the right side of the patterns `(x - h)^2 = x^2-2hx+h^2` and `(y-k)^2=y^2-2ky+k^2` and the terms in the parenthesis in equation [1] to find the values of h and k:

`x^2-2hx+h^2 =x^2+18x+h^2` and `y^2-2ky+k^2= y^2+10y+k^2`

`-2hx =18x` and `-2ky= 10y`

`h =-9` and `k= -5`

Substitute the left side of the patterns into equation [1]:

`4(x - h)^2-(y-k)^2=-199+4h^2-k^2" [2]"`

Substitute the values for h and k into equation [2]:

`4(x - -9)^2-(y- -5)^2=-199+4(-9)^2-(-5)^2" [3]"`

Simplify the right side of equation [3]:

`4(x - -9)^2-(y- -5)^2=100" [4]"`

Divide both sides of the equation by 100:

`(x - -9)^2/25-(y- -5)^2/100=1" [5]"`

Write the denominators as squares:

`(x - -9)^2/5^2-(y- -5)^2/10^2=1" [6]"`

Equation [6] is the standard form for a hyperbola with a horizontal transverse axis.

Referring to the standard form, the vertices are at `(h-a,k) and (h+a,k)`

Substituting `a = 5, h = -9, and k = -5`

`(-9-5,-5) and (-9+5,-6)`

The vertices are `(-16,-5) and (-4,-5)`

Referring to the standard form, the foci are at `(h-c,k) and (h+c,k)`

Use `c = sqrt(a^2+b^2)` with a = 5 and b = 10:

`c = sqrt(125)`

The foci are `(-9-sqrt(125),-5) and (-9+sqrt(125),-5)`

Referring to the standard form, the asymptotes are at:

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

Substituting `b = 10, a = 5, h = -9, and k = -5`

The asymptotes are: `y = -2(x--9)-5 and y = 2(x--9)-5`