Write the equation of the ellipse centered at zero that goes through the point $$(2,1)$$ and has a minor axis of length $$4$$.
Development:
If the minor axis measures $$4$$, we have $$2b=4 \Rightarrow b=\dfrac{4}{2}=2$$. Replacing $$b=2$$ in the equation of the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ we obtain $$a$$: $$$\dfrac{x^2}{a^2}+\dfrac{y^2}{2^2}=1 \text{ and pass through the point } (x,y)=(2,1)\Rightarrow$$$ $$$\Rightarrow \dfrac{2^2}{a^2}+\dfrac{1^2}{2^2}=1 \Rightarrow \dfrac{4}{a^2}+\dfrac{1}{4}=1 \Rightarrow a=\dfrac{4}{\sqrt{3}}$$$ Then the equation is: $$$\dfrac{x^2}{\Big(\dfrac{4}{\sqrt{3}}\Big)^2}+\dfrac{y^2}{2^2}=1 \Rightarrow \dfrac{3x^2}{16}+\dfrac{y^2}{4}=1$$$
Solution:
$$\dfrac{3x^2}{16}+\dfrac{y^2}{4}=1$$