Chapter 9.2, Problem 26E

Solution

Write the standard form of the equation of the ellipse with the given characteristic and center $\left(0,0\right)$ .

  $\frac{x^2}{25}+\frac{y^2}{16}=1$

Given information:

Vertex at: $\left(-5,0\right)$ , and focus at: $\left(3,0\right)$ .

Formula used:

• If the coordinates of the vertices are $\left(\pm a,0\right)$ and focus is $\left(\pm c,0\right)$ , then the major axis is parallel to x axis. Then use the equation is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

• If the coordinates of the vertices are $\left(0,\pm a\right)$ and focus is $\left(0,\pm c\right)$ , then the major axis is parallel to y axis. Then use the equation

  $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

• Using the equation $c^2=\left(a^2-b^2\right)$

Calculation:

In order to write the standard form of the equation of the ellipse with center at $\left(0,0\right)$ , vertex at $\left(-5,0\right)$ , and focus at $\left(3,0\right)$ , use the standard form of the equation of an ellipse with vertices as $\left(\pm a,0\right)$ and focus $\left(\pm c,0\right)$ , that is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

First find co-vertex ‘ b’ by using equation

  $c^2=\left(a^2-b^2\right)$

Thus, $a=-5\text{ and c}=3$ putting these values in above equation gives:

  $\begin{array}{l}{3}^2=\left({\left(-5\right)}^2-b^2\right)\\ 9=25-b^2\\ b^2+9=25\\ b^2=25-9\\ b^2=16\\ b=\sqrt{16}\\ b=4\end{array}$

Thus, $a=-5\text{ and }b=4$ putting these values in above equation gives:

  $\begin{array}{c}\frac{x^2}{{\left(-5\right)}^2}+\frac{y^2}{4^2}=1\\ \frac{x^2}{25}+\frac{y^2}{16}=1\end{array}$

Thus, the standard form of the equation of ellipse with given information is:

  $\frac{x^2}{25}+\frac{y^2}{16}=1$