Find an equation in standard form for the ellipse with center at $(6,5)$, horizontal axis of length 6 and passes through the point $(8,4)$.
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Step 1 ：The standard form of the equation of an ellipse with center at \((h, k)\), horizontal axis of length \(2a\) and vertical axis of length \(2b\) is given by: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

Step 2 ：Given that the center of the ellipse is at \((6,5)\), we have \(h = 6\) and \(k = 5\).

Step 3 ：The length of the horizontal axis is given as 6, so \(a = 6/2 = 3\).

Step 4 ：The ellipse passes through the point \((8,4)\), so we can substitute \(x = 8\) and \(y = 4\) into the equation of the ellipse to solve for \(b\).

Step 5 ：Substitute \(x = 8\), \(y = 4\), \(h = 6\), \(k = 5\), and \(a = 3\) into the equation of the ellipse: \[\frac{(8-6)^2}{3^2} + \frac{(4-5)^2}{b^2} = 1\]

Step 6 ：Solving this equation for \(b^2\) gives: \[\frac{4}{9} + \frac{1}{b^2} = 1\]

Step 7 ：Subtract \(\frac{4}{9}\) from both sides: \[\frac{1}{b^2} = 1 - \frac{4}{9} = \frac{5}{9}\]

Step 8 ：So, \(b^2 = \frac{9}{5}\).

Step 9 ：Therefore, the equation of the ellipse in standard form is: \[\frac{(x-6)^2}{3^2} + \frac{(y-5)^2}{(9/5)} = 1\]

Step 10 ：Or, simplifying further: \[\frac{(x-6)^2}{9} + \frac{5(y-5)^2}{9} = 1\]

Step 11 ：This is the equation of the ellipse in standard form. The final answer is \[\boxed{\frac{(x-6)^2}{9} + \frac{5(y-5)^2}{9} = 1}\]