Chapter 9, Problem 26CR

To determine

To find: the standard form of equation of ellipse.

Answer to Problem 26CR

The standard form of ellipse is $\frac{{\left(x-2\right)}^2}{4}+\frac{{\left(y-2\right)}^2}{4}=1$ .

Explanation of Solution

Given information:

Vertex: $\left(2,0\right),\left(2,4\right)$ , foci: $\left(2,1\right),\left(2,3\right)$

Calculation: to find the standard form of ellipse,

The centre of the ellipse is the midpoint of the line segment joining the foci.

So by the midpoint formula, the centre of the ellipse is $\left(2,2\right)$ .

The distance from the centre to one foci is $c=0$ .

Since, the vertex of the ellipse is $\left(2,0\right),\left(2,4\right)$ , so the value of $a$ is $a=2$ .

Now, by using the relation between $a,b,c$ ,

  $\begin{array}{l}{c}^2=a^2-b^2\\ {\left(0\right)}^2={\left(2\right)}^2-b^2\text{ [substitute]}\\ b^2=4-0\text{ [simplify]}\\ b^2=4\text{ [subtract]}\end{array}$

Since, the standard form of the ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ ,

Now, substitute and simplify,

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-2\right)}^2}{4}+\frac{{\left(y-2\right)}^2}{4}=1\text{ [substitute]}\end{array}$

Thus, The standard form of ellipse is $\frac{{\left(x-2\right)}^2}{4}+\frac{{\left(y-2\right)}^2}{4}=1$ .