SOLUTION: Find the equation (hyperbola) if the asymptotes y - 3 = √13/6 (x-2) and focus at (9, 3)

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Question 1197351: Find the equation (hyperbola) if the
asymptotes y - 3 = √13/6 (x-2) and focus at (9, 3)

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If you use the "√" symbol for a square root, then use parentheses to make it clear exactly what part of the expression is under the radical. "√13/6" could mean either sqrt(13)/6 or sqrt(13/6).

A coefficient like sqrt(13/6) would probably not be given in a problem, since it is not in simplified form; so I will assume the coefficient is sqrt(13)/6.

You show the equation of one of the asymptotes; I further assume that the asymptotes (plural) have equations with coefficients sqrt(13)/6 and -sqrt(13)/6.

The equations of the asymptotes tell us the center of the hyperbola is (2,3); and the focus at (9,3) tells us the branches of the hyperbola open right and left. So we know the equation of the hyperbola has the form

%28x-2%29%5E2%2Fa%5E2-%28y-3%29%5E2%2Fb%5E2=1

The slopes of the asymptotes tell us that b%2Fa=sqrt%2813%29%2F6.

Finally, with the center at (2,3) and one focus at (9,3), we know that c, the distance from the center to each focus, is 7.

So we have...
b/a = sqrt(13)/6;
c = 7;
and, for a hyperbola, c%5E2+=+a%5E2%2Bb%5E2

Then the simple observation that 7%5E2+=+sqrt%2813%29%5E2%2B6%5E2 tells us that b = sqrt(13) and a = 6.

So the equation of the hyperbola is

%28x-2%29%5E2%2F36-%28y-3%29%5E2%2F13=1