Problem 1
Answer: Choice D
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Explanation:
The vertices are (-12,0) and (12,0). Both of which are on the same horizontal line. Because of this, we have a hyperbola that opens leftward and rightward. Such hyperbolas are of the form
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
where (h,k) is the center and the a,b terms help set up the asymptotes.
The midpoint of the vertices (-12,0) and (12,0) is the point (0,0). This is the center. So we know that h = 0 and k = 0.
For left/right opening hyperbolas, the asymptotes can be described as the equations:
y = (b/a)x
y = (-b/a)x
Or we can condense them into [tex]y = \pm \frac{b}{a}x[/tex]. Compare this to the given equation [tex]y = \pm\frac{5}{12}x[/tex] to see that b = 5 and a = 12.
Therefore, we get the following:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\\\\\frac{(x-0)^2}{12^2}-\frac{(y-0)^2}{5^2}=1\\\\\frac{x^2}{144}-\frac{y^2}{25}=1[/tex]
That points us to choice D.
You chose the correct answer.
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Problem 2
Answer: Choice A
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Explanation:
Divide both sides by 3600 so we can turn that "3600" into "1". Then our goal is to get that "1" to its own side like so:
[tex]25x^2-144y^2 + 3600 = 0\\\\\frac{25x^2-144y^2 + 3600}{3600} = \frac{0}{3600}\\\\\frac{25x^2}{3600}-\frac{144y^2}{3600} + \frac{3600}{3600} = 0\\\\\frac{x^2}{144}-\frac{y^2}{25} + 1 = 0\\\\\frac{x^2}{144}-\frac{y^2}{25} = -1\\\\-\frac{x^2}{144}+\frac{y^2}{25} = 1\\\\-\frac{x^2}{12^2}+\frac{y^2}{5^2} = 1\\\\[/tex]
We see that a = 12 and b = 5 like earlier, so we end up with the same asymptotes as before. The only difference is that this hyperbola opens upward/downward rather than left/right.
You made the correct choice in picking answer A. Both of your answers are correct. Nice work.
