Problems from Equation of the ellipse with foci on the axis OX

Development:

If the minor axis measures $4$, we have $2b=4\Rightarrow b={\displaystyle \frac{4}{2}}=2$. Replacing $b=2$ in the equation of the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ we obtain $a$: $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{2^2}}=1\text{ and pass through the point }(x,y)=(2,1)\Rightarrow$$ $$\Rightarrow {\displaystyle \frac{2^2}{a^2}}+{\displaystyle \frac{1^2}{2^2}}=1\Rightarrow {\displaystyle \frac{4}{a^2}}+{\displaystyle \frac{1}{4}}=1\Rightarrow a={\displaystyle \frac{4}{\sqrt{3}}}$$ Then the equation is: $${\displaystyle \frac{x^2}{({\displaystyle \frac{4}{\sqrt{3}}}{)}^2}}+{\displaystyle \frac{y^2}{2^2}}=1\Rightarrow {\displaystyle \frac{3x^2}{16}}+{\displaystyle \frac{y^2}{4}}=1$$

Solution:

${\displaystyle \frac{3x^2}{16}}+{\displaystyle \frac{y^2}{4}}=1$

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Development:

Since the foci are in the axis $OX$ we must use Equation I of the ellipse $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$

It is necessary for us to determine the value of the semiaxes $a$ and $b$, which are the major semiaxis and the minor semiaxis respectively.

The statement tells us that the ellipse passes through point therefore it determined at which the moment in which the ellipse cuts the the axis of $x$, since the value of $y$ at this point it is zero.

And so, in fact it shows us to what the distance is from the center of the ellipse to the cutting point between this and the axis $OX$, which we have defined as the major semiaxis.

Therefore, the value of the biggest semiaxis is $3$. Namely, $a=3$.

Since we also know the distance $c$ from the center to the focus (which is $2$), in viture of the relation $a^2=b^2+c^2$ we can isolate $b$ and find: $$b=\sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5}$$ Once we know all the parameters of the ellipse, write its equation: $${\displaystyle \frac{x^2}{3^2}}+{\displaystyle \frac{y^2}{(\sqrt{5}{)}^2}}=1$$ $${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{5}}=1$$

If we want to calculate its eccentricity it is only necessary to divide $c$ and $a$. That is: $$e={\displaystyle \frac{2}{3}}$$

Solution:

${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{5}}=1$, $e={\displaystyle \frac{2}{3}}$

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