Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 637: 13

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Given: $16x^2+25y^2=1600\\\frac{x^2}{100}+\frac{y^2}{64}=1$ The equation is in standard form. We can see $a=10, b=8$ The denominator of the $x^2-term$ is greater than that of the $y^2-term$, so the major axis is horizontal. The vertices of the ellipse are at $(\pm a,0)=(\pm 10,0)$. The co-vertices are at $(0,\pm b) = (0,\pm 8)$. Find the foci. $c^2=a^2-b^2=10^2-8^2=36$ so $c=6$ The foci are at $(\pm 6,0)$.