Chapter 8.2, Problem 49E

Solution

An equation for the given ellipse.

It has been determined that the required equation of the given ellipse is $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-3\right)}^2}{9}=1$ .

Given:

The following figure of an ellipse:

  

Concept used:

The equation of an ellipse in standard form, having center $\left(h,k\right)$ , focal axis $y=k$ , foci $\left(h\pm c,k\right)$ , vertices $\left(h\pm a,k\right)$ , semi-major axis $a$ and semi-minor axis $b$ ; is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ , where $c^2=a^2-b^2$ .

The equation of an ellipse in standard form, having center $\left(h,k\right)$ , focal axis $x=h$ , foci $\left(h,k\pm c\right)$ , vertices $\left(h,k\pm a\right)$ , semi-major axis $a$ and semi-minor axis $b$ ; is $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$ , where $c^2=a^2-b^2$ .

Calculation:

It can be seen from the given figure that the ellipse has its center at $\left(2,3\right)$ , a vertex at $\left(6,3\right)$ and one end point of the minor axis at $\left(2,6\right)$ .

Comparing the given center $\left(2,3\right)$ with $\left(h,k\right)$ , the given vertex $\left(6,3\right)$ with $\left(h\pm a,k\right)$ and the given end point of the minor axis $\left(2,6\right)$ with $\left(h,k\pm b\right)$ , it follows that:

  $h=2$ , $k=3$ , $h+a=6$ and $k+b=6$ .

Put $h=2$ in $h+a=6$ to get $a=4$ .

Put $k=3$ in $k+b=6$ to get $b=3$ .

Put $h=2$ , $k=3$ , $a=4$ and $b=3$ in $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ to get,

  $\frac{{\left(x-2\right)}^2}{4^2}+\frac{{\left(y-3\right)}^2}{3^2}=1$

Simplifying,

  $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-3\right)}^2}{9}=1$

This is the required equation of the given ellipse.

Conclusion:

It has been determined that the required equation of the given ellipse is $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-3\right)}^2}{9}=1$ .