Respuesta :
so hmm if you check the picture below
those are the given foci and co-vertices
so, since the co-vertices are at 2,0 and -2,0 that makes the center, at the origin and the "b" component 2, so the minor axis is just 2+2 or 4
now, what's the major axis? or what is the "a" component?
based on the provided points, is a vertical ellipse, that means the major axis runs over the y-axis and thus the "a" or larger denominator, lies under the fraction with the "y"
now
[tex]\bf \textit{ellipse, vertical major axis}\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}, {{ k}}\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{{{ a }}^2-{{ b }}^2}\\ ----------\\ h=0\\ k=0\\ b=2\\ c=6 \end{cases}[/tex]
[tex]\bf c=\sqrt{a^2-b^2}\implies 6=\sqrt{a^2-4}\implies 36=a^2-4 \\\\\\ \boxed{40=a^2}\implies \sqrt{40}=a\implies 2\sqrt{10}=a\\\\ -----------------------------\\\\ \cfrac{(x-0)^2}{4}+\cfrac{(y-0)^2}{a^2}=1\implies \cfrac{x^2}{4}+\cfrac{y^2}{40}=1[/tex]
those are the given foci and co-vertices
so, since the co-vertices are at 2,0 and -2,0 that makes the center, at the origin and the "b" component 2, so the minor axis is just 2+2 or 4
now, what's the major axis? or what is the "a" component?
based on the provided points, is a vertical ellipse, that means the major axis runs over the y-axis and thus the "a" or larger denominator, lies under the fraction with the "y"
now
[tex]\bf \textit{ellipse, vertical major axis}\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}, {{ k}}\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{{{ a }}^2-{{ b }}^2}\\ ----------\\ h=0\\ k=0\\ b=2\\ c=6 \end{cases}[/tex]
[tex]\bf c=\sqrt{a^2-b^2}\implies 6=\sqrt{a^2-4}\implies 36=a^2-4 \\\\\\ \boxed{40=a^2}\implies \sqrt{40}=a\implies 2\sqrt{10}=a\\\\ -----------------------------\\\\ \cfrac{(x-0)^2}{4}+\cfrac{(y-0)^2}{a^2}=1\implies \cfrac{x^2}{4}+\cfrac{y^2}{40}=1[/tex]
