Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 637: 8

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Given: $\frac{x^2}{36}+\frac{y^2}{225}=1$ The equation is in standard form. We can see $a=15, b=6$ 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 $(0,\pm a)=(0,\pm 15)$. The co-vertices are at $(\pm b,0) = (\pm 6,0)$. Find the foci. $c^2=a^2-b^2=15^2-6^2=189$ so $c=3\sqrt 21$ The foci are at $(0,\pm 3\sqrt 21)$.