Given the equation of an ellipse, (frac{(x-5)^2}{16} + frac{(y+3)^2}{9} = 1), find the vertex form of the ellipse.

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Accepted Answer

Step:1 ：First, observe that the given equation is already in the standard form of an ellipse with center at (h, k), which is (frac{(x-h)^2}{a^2} + frac{(y-k)^2}{b^2} = 1). Comparing, we find h = 5, k = -3, (a^2 = 16) and (b^2 = 9).Step:2 ：Next, find the values of a and b. Since (a^2 = 16) and (b^2 = 9), we have a = 4 and b = 3.Step:3 ：The vertices of an ellipse in the standard form are at (h±a, k) and (h, k±b). So, substituting the values of h, k, a and b, we find the vertices at (5±4, -3) and (5, -3±3).

Answer

Step: 1 ：First, observe that the given equation is already in the standard form of an ellipse with center at (h, k), which is (frac{(x-h)^2}{a^2} + frac{(y-k)^2}{b^2} = 1). Comparing, we find h = 5, k = -3, (a^2 = 16) and (b^2 = 9).Step: 2 ：Next, find the values of a and b. Since (a^2 = 16) and (b^2 = 9), we have a = 4 and b = 3.Step: 3 ：The vertices of an ellipse in the standard form are at (h±a, k) and (h, k±b). So, substituting the values of h, k, a and b, we find the vertices at (5±4, -3) and (5, -3±3).

Given the equation of an ellipse, \(\frac{(x-5)^2}{16} + \frac{(y+3)^2}{9} = 1\), find the vertex form of the ellipse.

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