Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 638: 42

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Given: $\frac{3x^2}{48}+\frac{4y^2}{400}=1\\\frac{x^2}{16}+\frac{y^2}{100}=1$ The equation is in standard form. We can see $a=10, b=4$. The denominator of the $x^2-term$ is smaller than that of the $y^2-term$, so the major axis is vertical. The vertices of the ellipse are at $(0,\pm a)=(0,\pm 10)$. The co-vertices are at $(\pm b,0) = (\pm 4,0)$.