How do you find the vertices and foci of #y^2/49-x^2=1#?

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Answer 1 · 2017-01-26T03:10:10

Given the general Cartesian form:

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

The vertices are #(h, k - a) and (h, k + a)#

The foci are #(h, k - sqrt(a^2 + b^2)) and (h, k + sqrt(a^2 + b^2))#

Write the given equation, #y^2/49 - x^2 = 1#, in the form of equation [1]:

Insert -0 with the square both numerators:

#(y-0)^2/49 - (x-0)^2 = 1#

Write the 49 as #7^2# and write a #1^2# under the x term:

#(y - 0)^2/7^2 - (x - 0)^2/1^2 = 1" [2]"#

Matching the variables in equation [1] with the values in equation [2], #h=0,k=0,a=7, and b=1#

Compute #sqrt(a^2 + b^2) = sqrt(7^2 + 1^2) = sqrt(50) = 5sqrt(2)#

Now, it is a simple matter to write the vertices and foci.

Vertices: #(0, -7) and (0, 7)#

Foci: #(0, -5sqrt2) and (0,5sqrt2)#