Chapter 12, Problem 1RE

For the following exercises. write the equation of the ellipse in standard form. Then identify the center, vertices, and foci.

1. $\frac{x^2}{25}+\frac{y^2}{64}=1$

To determine

To calculate: To write the equation in standard form and to state the center, vertices and foci.

Answer to Problem 1RE

The center of the ellipse is $\left(0,0\right)$

The coordinates of the vertices are $\left(0,\pm 8\right)$

The coordinates of the foci are $\left(0,\pm \sqrt{39}\right)$

Explanation of Solution

Given information:

Given, equation is $\frac{x^2}{25}+\frac{y^2}{64}=1$

Calculation:

The standard form of equation of an ellipse with major axis along x -axis i.e. $a>b$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The standard form of equation of an ellipse with major axis along y -axis i.e. $a>b$ is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Here $64>25$ .

Hence the major axis is along the y -axis.

Therefore, $a^2=64,b^2=25$

For an ellipse, $c^2=a^2-b^2=64-25=39$

  $\Rightarrow a=8,b=5,c=\sqrt{39}$

When an ellipse is of the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ ,

The center of the ellipse is $\left(0,0\right)$

The coordinates of the vertices are $\left(0,\pm a\right)=\left(0,\pm 8\right)$

The coordinates of the foci are $\left(0,\pm c\right)=\left(0,\pm \sqrt{39}\right)$