Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 638: 44

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Given: $16x^2+10y^2=160\\\frac{x^2}{10}+\frac{y^2}{16}=1$ The equation is in standard form. We can see $a=4, b=\pm \sqrt 10$ 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 $(0,\pm a)=(0,\pm 4)$. The co-vertices are at $(\pm b,0) = (\pm \sqrt 10,0)$.