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

1 Answer
Mar 15, 2017

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#