Chapter 13.6, Problem 46E

To determine

To find: the foci, center and vertices of the ellipse with the equation given below.

Answer to Problem 46E

The foci here is $\left(2\sqrt{6},-3\right),\left(-2\sqrt{6},-3\right)$ ,vertices along the major axis is $\left(7,-3\right),\left(-7,-3\right)$ and vertices along the minor axis is $\left(0,2\right),\left(0,-8\right)$ and Center is $\left(0,-3\right)$ ..

Explanation of Solution

Given:

  $\frac{x^2}{49}+\frac{{\left(y+3\right)}^2}{25}=1$ .

Concept used:

Major axis is $2a$ and $a$ will be chosen normally higher value from the ellipse equation.

Similarly, minor axis is $2b$ and $b$ will be chosen normally lower value from the ellipse equation.

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Focal distance:

  $c^2=a^2-b^2$ .

Calculation:

Here from the given equation:

  $\frac{x^2}{49}+\frac{{\left(y+3\right)}^2}{25}=1$ .

Center: $\left(0,-3\right)$

  $a=7$ ,so:

Vertices along the major axis:

  $\begin{array}{l}\left(7,-3\right),\left(-7,-3\right)\\ b=5.\end{array}$

So, vertices along the minor axis:

  $\left(0,2\right),\left(0,-8\right)$ .

Focal distance:

  $c=\sqrt{49-25}=\sqrt{24}=2\sqrt{6}.$

Foci:

  $\left(2\sqrt{6},-3\right),\left(-2\sqrt{6},-3\right)$ .

Hence, the foci here is $\left(2\sqrt{6},-3\right),\left(-2\sqrt{6},-3\right)$ ,vertices along the major axis is $\left(7,-3\right),\left(-7,-3\right)$ and vertices along the minor axis is $\left(0,2\right),\left(0,-8\right)$ and Center is $\left(0,-3\right)$ .