Chapter 9.4, Problem 18MCQ

To determine

To find: The coordinates of the centre and foci and the length of the major and minor axis for the given ellipse.

Answer to Problem 18MCQ

The coordinates of the centre and foci are $\left(1,-2\right)$ and $\left(-3,-2\right),\left(5,-2\right)$ , the length of the major axis is $4\sqrt{5}\text{unit}$ , and the length of the minor axis is $4\text{unit}$ .

Explanation of Solution

Given information:

The equation of the ellipse is,

  $\frac{{\left(x-1\right)}^2}{20}+\frac{{\left(y+2\right)}^2}{4}=1$

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

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

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

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

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

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

Compare the given equation with the standard form of the equation to obtain,

  $\begin{array}{l}h=1\\ k=-2\\ a^2=20\\ a=2\sqrt{5}\end{array}$

  $\begin{array}{l}{b}^2=4\\ b=2\end{array}$

The length of the major axis is calculated as,

  $\begin{array}{l}2a=2\times 2\sqrt{5}\\ 2a=4\sqrt{5}\text{unit}\end{array}$

The length of the minor axis is calculated as,

  $\begin{array}{l}2b=2\times 2\\ 2b=4\text{unit}\end{array}$

The value of $c$ is calculated as,

  $\begin{array}{c}c=\sqrt{a^2-b^2}\\ =\sqrt{20-4}\\ =4\text{unit}\end{array}$

The coordinate of the foci is calculated as,

  $\begin{array}{c}\left(h\pm c,k\right)=\left(1\pm 4,-2\right)\\ =\left(-3,-2\right),\left(5,-2\right)\end{array}$

Consider the following graph of the ellipse,

  

Thus, the coordinates of the centre and foci are $\left(1,-2\right)$ and $\left(-3,-2\right),\left(5,-2\right)$ , the length of the major axis is $4\sqrt{5}\text{unit}$ , and the length of the minor axis is $4\text{unit}$ .