Chapter 8, Problem 44RE

Solution

To find: the equation for the conic in the standard form.

The standard equation of the ellipse with center $\left(-3,-4\right)$ and focal axis $y=-4$ is $\frac{{\left(x+3\right)}^2}{16}+\frac{{\left(y+4\right)}^2}{7}=1$ .

Given data:

Type of conic: Ellipse

Center of the ellipse $=\left(-3,-4\right)$

Semi-major axis $=4$

One focus is $\left(0,-4\right)$ .

Formula used:

The standard equation of the ellipse with center $\left(h,k\right)$ and focal axis $y=k$ is given as:

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

Calculation:

It is given that the center of the ellipse is $\left(-3,-4\right)$ and semi-major axis is $4$ .

  $h=-3,k=-4,a=4$

Foci of the ellipse is given as $\left(h\pm c,k\right)$ . One of the focus is $\left(0,-4\right)$ .

  $\begin{array}{c}h\pm c=0\\ -3\pm c=0\\ \pm c=3\end{array}$

Calculate $b$ using Pythagorean theorem.

  $\begin{array}{c}{a}^2=b^2+c^2\\ 4^2=b^2+3^2\\ b^2=\left(4-3\right)\left(4+3\right)\\ b^2=7\\ b=\pm \sqrt{7}\end{array}$

The standard equation of the ellipse with center $\left(-3,-4\right)$ and focal axis $y=-4$ is given as:

  $\begin{array}{c}\frac{{\left(x-\left(-3\right)\right)}^2}{4^2}+\frac{{\left(y-\left(-4\right)\right)}^2}{{\left(\sqrt{7}\right)}^2}=1\\ \frac{{\left(x+3\right)}^2}{16}+\frac{{\left(y+4\right)}^2}{7}=1\end{array}$

Therefore, the standard equation of the ellipse with center $\left(-3,-4\right)$ and focal axis $y=-4$ is $\frac{{\left(x+3\right)}^2}{16}+\frac{{\left(y+4\right)}^2}{7}=1$ .