91Ó°ÊÓ

Center at $(1,2)$: vertex at $(4,2)$:contains the point $(1,5)$.

On comparing with the standard equation of the ellipse $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$

Here,$x=1,y=5,h=1,k=2$and $$\begin{gathered} h+a=4 \\ a=4-1 \\ a=3 \end{gathered}$$

Now put these values in the equation of the ellipse.

$$\begin{gathered} \frac{{(1-1)}^2}{3^2}+\frac{{(5-2)}^2}{b^2}=1 \\ \frac{0}{9}+\frac{9}{b^2}=1 \\ \frac{9}{b^2}=1 \\ {b}^2=9 \end{gathered}$$

Substitute the value of $a^2=9$and $b^2=9$we get

$\frac{{(x-1)}^2}{9}+\frac{{(y-2)}^2}{9}=1$

The major axis is parallel to the x-axis.So the vertices are $a=3$units left and right of the center $(1,2)$.Therefore the vertices are localid="1646732539408" $v_1=(-2,2)$and localid="1646732554304" $v_2=(4,2)$

Since,

localid="1646732713719" $$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=9-9 \\ {c}^2=0 \end{gathered}$$

Major axis is parallel to the x-axis, foci are localid="1646732722940" $(1-0,2)$and localid="1646732737337" $(1,2)$.

We use the value localid="1646732752572" $b=3$to find the two points above and below the center aslocalid="1646732764537" $(1,-1)$andlocalid="1646732779266" $(1,5)$