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How do you identify the vertices, foci, and direction of #(x-3)^2/4-(y+1)^2/9=1#?

Answer 1

The direction is horizontal transverse
The vertices are at #(1, -1)# and #(5, -1)#
The foci are #(3 - sqrt(13), -1)# and #(3 + sqrt(13), -1)#

Please notice that the center of the hyperbola is #(3, -1)#.

To find the vertices, force the negative term to become zero by setting #y = -1#:

#((x - 3)²)/4 = 1#

#(x - 3)² = 4#

We are about to take the square root of both sides so we need to write two equation one where the right side is 2 and the other where the right side is -2:

#x - 3 = 2# and #x - 3 = -2#

#x = 5, and x = 1#

The vertices are at #(1, -1)# and #(5, -1)#

We know that the y coordinate of both foci is -1 but the x coordinate is plus or minus the distance c from the x coordinate of the center, 3 The formula for c is:

#c² = a² + b²# where #a²# and #b²# are the denominators in the equation.

#c² = 4 + 9#

#c = sqrt(13)#

The foci are #(3 - sqrt(13), -1)# and #(3 + sqrt(13), -1)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.