Question

What are the vertices of the ellipse given by the equation x2 + 4y2 + 10x – 56y + 205 = 0

Answer 1

Complete the square... the vertex form of an ellipse is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1\), where the vertex is at \((h,k)\).

Answer 2

you got this?

Answer 3

\[x^2 + 4y^2 + 10x – 56y + 205 = 0\]
\[(x+5)^2-4(y-7)^2=-205+25+196=16\]
\[\frac{(x+5)^2}{16}-\frac{(y-7)^2}{4}=1\]

Answer 4

center is \((-5,7)\) so the vertices are 4 units to the left and right at \((-9,7)\) and \((-1,7)\)