Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 637: 3

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Given: $\frac{x^2}{16}+\frac{y^2}{4}=1$ The equation is in standard form. We can see $a=4, b=2$ 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 $(\pm a,0)=(\pm 4,0)$. The co-vertices are at $(0, \pm b) = (0, \pm 2)$. Find the foci. $c^2=a^2-b^2=4^2-2^2=12$ So $c=2\sqrt 3$ The foci are at $(\pm 2\sqrt 3 , 0)$.