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

1 Answer
Jun 25, 2017

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))#