Chapter 12, Problem 3CLT

To determine

To find: The standard equation of the ellipse with given characteristics.

Answer to Problem 3CLT

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

Explanation of Solution

Given: Vertices of the ellipse are (0,0) and (0,4); endpoints of minor axis are (1,2) and (-1,2).

Calculation:

Let a and b be length of semi major axis and semi minor axis respectively.

Here, we have ellipse with major axis on y -axis(since both the vertices are on line $x=0$ )

Length of major axis =4-0=4

Therefore,

  $\begin{array}{l}2a=4\\ a=2\end{array}$

The minor axis is in line $y=2$ (since both the end-points of minor axis are on line $y=2$ )

Length of minor axis 1-(-1)=2

Therefore,

  $\begin{array}{l}2b=2\\ b=1\end{array}$

Now, center of ellipse is mid-point of the line segment joining vertices of ellipse, so center of ellipse is $\left(\frac{0+0}{2},\frac{0+4}{2}\right)$ , that is $\left(0,2\right)$

We know that the standard equation of the vertical ellipse with center at ( h , k ) is

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

Therefore, standard equation of ellipse with given characteristics is

  $\begin{array}{l}\frac{{\left(y-2\right)}^2}{2^2}+\frac{{\left(x-0\right)}^2}{1^2}=1\\ \frac{{\left(y-2\right)}^2}{2^2}+\frac{x^2}{1^2}=1\end{array}$