SOLUTION: Foci at (-1,4) and (7,4), transverse axis has the length 8/3. Whats the equation

Question 1168978: Foci at (-1,4) and (7,4), transverse axis has the length 8/3. Whats the equation
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Solution:
The given information indicates that the conic section is a hyperbola because we are given foci and a transverse axis length.
The foci are at $F_1(-1, 4)$ and $F_2(7, 4)$.
The transverse axis has a length of $\frac{8}{3}$.
From the foci, we can determine the center of the hyperbola as the midpoint of the foci:
Center $(h, k) = \left( \frac{-1 + 7}{2}, \frac{4 + 4}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)$.
The distance between the foci is $2c$.
$2c = \sqrt{(7 - (-1))^2 + (4 - 4)^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8$.
So, $c = 4$.
The length of the transverse axis is $2a$. We are given $2a = \frac{8}{3}$, so $a = \frac{4}{3}$.
Since the y-coordinates of the foci are the same, the transverse axis is horizontal. The standard form of the equation of a hyperbola with a horizontal transverse axis and center $(h, k)$ is:
$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$
We need to find $b^2$. We use the relationship $c^2 = a^2 + b^2$ for a hyperbola.
$4^2 = \left(\frac{4}{3}\right)^2 + b^2$
$16 = \frac{16}{9} + b^2$
$b^2 = 16 - \frac{16}{9} = \frac{144 - 16}{9} = \frac{128}{9}$.
Now, substitute the values of $h=3$, $k=4$, $a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$, and $b^2 = \frac{128}{9}$ into the standard form:
$\frac{(x - 3)^2}{\frac{16}{9}} - \frac{(y - 4)^2}{\frac{128}{9}} = 1$
Multiply the numerators by $\frac{9}{16}$ and $\frac{9}{128}$ respectively:
$\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1$.
Final Answer: The final answer is $\boxed{\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1}$
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