Question

Find an equation for each ellipse. Graph the equation by hand.

Center at$(1,2)$vertex at$(1,4)$ contains the point$(1+\sqrt{3},3)$.

Short answer

The equation of the ellipse is $\frac{{(x-1)}^2}{4}+\frac{{(y-2)}^2}{4}=1$and graph of the ellipse is

Step-by-step solution

. Given information. 

Center at $(1,2)$vertex at $(1,4)$contains the points $(1+\sqrt{3},3)$.

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

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

Step 2. The equation of ellipse. 

Substituting the values in the equation of ellipse.

$$\begin{gathered} \frac{{(1+\sqrt{3}-1)}^2}{2^2}+\frac{{(3-2)}^2}{b^2}=1 \\ \frac{{\left(\sqrt{3}\right)}^2}{4}+\frac{1}{b^2}=1 \\ \frac{1}{b^2}=1-\frac{3}{4} \\ \frac{1}{b^2}=\frac{1}{4} \\ {b}^2=4 \end{gathered}$$

Thus, the equation of the ellipse is$\frac{{(x-1)}^2}{4}+\frac{{(y-2)}^2}{4}=1$

Step 3. Graph of the ellipse .   

The major axis is parallel to the y-axis. so the vertices are $a=2$units up and down of the center $(1,2)$. Therefore the vertices are $v_1=(1,0)$and $v_2=(1,4)$.

Since,

$$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=4-4 \\ {c}^2=0 \\ c=0 \end{gathered}$$

The major axis is parallel to the y-axis foci are $(1,2-0)$and $(1,2+0)$or $(1,2)$.

Finally, we use the value $b=2,b=3$to find the two points above and below the center as $(-1,2)$and $(3,2)$.