Question
In Exercises $37-50,$ graph each ellipse and give the location of its foci.$$\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1$$
Step 1
Comparing this with the given equation, we get $h=3$, $k=-1$, $a^{2}=9$ and $b^{2}=16$. So, $a=3$ and $b=4$. Show more…
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In Exercises $37-50,$ graph each ellipse and give the location of its foci. $$ \frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1 $$
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In Exercises $37-50,$ graph each ellipse and give the location of its foci. $$ \frac{(x+3)^{2}}{9}+(y-2)^{2}=1 $$
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