Problems from Equation of the ellipse with center (x0, y0) and focal axis parallel to y axis

Development:

a) For this case, since it is centred on the zero and the focus is in the axis $OX$, we use the first equation of the ellipse.

We see that the major semiaxis measures $3$ and the equation is: $${\displaystyle \frac{x^2}{3^2}}+{\displaystyle \frac{y^2}{b^2}}=1\Rightarrow \text{ since }c=2\text{ we obtain }{b}^2=3^2-2^2=5\Rightarrow b=\sqrt{5}$$

The equation will be ${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{5}}=1$.

b) For this case, since it is not centred on the zero and given that it has the focus in the axis that is parallel to the $OY$, we use the formula. We also know that $b=4$ and $c=3$ therefore $a$ is: $a=\sqrt{16+9}=5$. Then the equation is: $${\displaystyle \frac{(y+1{)}^2}{25}}+{\displaystyle \frac{(x-1{)}^2}{16}}=1$$

Solution:

a) The equation is: ${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{5}}=1$

b) The equation is: ${\displaystyle \frac{(y+1{)}^2}{25}}+{\displaystyle \frac{(x-1{)}^2}{16}}=1$

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