ANSWER:
[tex]\begin{gathered} \frac{(x+5)^2}{5^2}+\frac{(y-2)^2}{2^2}=1 \\ \text{end points of the major axis }=(0,2),(-10,2) \\ \text{end points of the minor axis }=(-5,4),(-5,0) \\ f=(-5+\sqrt[]{21},2),(-5-\sqrt[]{21},2) \end{gathered}[/tex]STEP-BY-STEP EXPLANATION:
We have the following equation of the ellipse in its standard form:
[tex]4x^2+40x+25y^2-100y+100=0[/tex]We convert this equation to the general form of an equation of an ellipse with no center at the origin, like this:
[tex]\begin{gathered} 4x^2+40x+25y^2-100y=-100 \\ 4\cdot(x^2+10x)+25\cdot(y^2-4y)=-100 \\ \text{ We divide by 4} \\ (x^2+10x)+\frac{25}{4}\cdot(y^2-4y)=-25 \\ \text{ We divide by 25} \\ \frac{1}{25}(x^2+10x)+\frac{1}{4}\cdot(y^2-4y)=-1 \\ \text{ We complete both squares} \\ \frac{1}{25}(x^2+10x+25)+\frac{1}{4}\cdot(y^2-4y+4)=-1+\frac{1}{25}\cdot25+\frac{1}{4}\cdot4 \\ \frac{1}{25}(x+5)^2+\frac{1}{4}(y-2)^2=-1+1+1 \\ \frac{(x+5)^2}{25}+\frac{(y-2)^2}{4}=1 \\ \frac{(x-(-5))^2}{5^2}+\frac{(y-2)^2}{2^2}=1 \end{gathered}[/tex]We have that the equation in its general form is the following:
[tex]\begin{gathered} \frac{\mleft(x-h\mright)^2}{a^2}+\frac{\mleft(y-k\mright)^2}{b^2}=1 \\ \text{ Therefore, in this case:} \\ h=-5 \\ k=2 \\ a=5 \\ b=2 \\ \text{end points of the major axis }=(-5+5,2),(-5-5,2)\rightarrow(0,2),(-10,2) \\ \text{end points of the minor axis }=(-5,2+2),(-5,2-2)\rightarrow(-5,4),(-5,0) \end{gathered}[/tex]Now, we calculate the focus of the ellipse just like this:
[tex]\begin{gathered} c^2=a^2-b^2 \\ c=\sqrt[]{5^2-2^2} \\ c=\sqrt[]{25-4} \\ c=\sqrt[]{21} \\ \text{ Therefore, the foci is:} \\ f=(-5+\sqrt[]{21},2),(-5-\sqrt[]{21},2) \end{gathered}[/tex]