Determine the center, vertices, foci, eccentricity and lengths of the major and minor axes of the ellipse $\displaystyle \frac{x^2}{49} + \frac{y^2}{9} = 1$. Then sketch its graph.
The equation has the form $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with horizontal major since the denominator of $x^2$ is larger. This gives $a^2 = 4a$ and $b^2 = 9$, so $c^2 = a^2 - b^2 = 49- 9 = 40$. Thus, $a = 7, b = 3$ and $c = \sqrt{40} = 2 \sqrt{10}$. Then, the following is determined as
vertices $(\pm a, 0) \to (\pm 7,0)$
foci $(\pm c, 0) \to (\pm 2 \sqrt{10}, 0)$
eccentricity $\displaystyle \frac{c}{a} \to \frac{2 \sqrt{10}}{7}$
length of the major axis $2a \to 14$
length of the minor axis $2b \to 6$
Therefore, the graph is
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