Chapter 10.3, Problem 63AYU

In Problems 55-64, find an equation for each ellipse. Graph the equation by hand.

Center at $\left(1,2\right)$; vertex at $\left(4,2\right)$; contains the point $\left(1,5\right)$

To determine

(a & b)

To find: An equation for each ellipse.

Answer to Problem 63AYU

Solution:

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

Explanation of Solution

Given:

Center at $\left(1,2\right)$; vertex at $\left(4,2\right)$; contains the point $\left(1,5\right)$.

Formula used:

Center	Major axis	Foci	Vertices	Equation
$\left(h, k\right)$	$\text{parallel to }x\text{-axis}$	$\left(h+c,k\right)$	$\left(h+a,k\right)$	$\begin{array}{l}\left(x-h\right)²/a²+\left(y-k\right)²/b²=1\hfill \\ a>b>0\text{ and }b²=a²-c²\hfill \end{array}$
$\left(h, k\right)$	$\text{parallel to }x\text{-axis}$	$\left(h-c,k\right)$	$\left(h-a,k\right)$

Calculation:

The center of the ellipse $\left(h,k\right)=\left(1,2\right)$.

$h=1;k=2$

Vertex: $\left(4,2\right)=\left(h+a,k\right);a=3$.

Major axis is parallel to $x\text{-axis}$.

Equation of the ellipse is given by,

$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b²}=1$

$\frac{{\left(x-1\right)}^2}{a²}+\frac{{\left(y-2\right)}^2}{b²}=1$

The equation contains the point $\left(1,5\right)$.

Substituting the points, we get,

$0+\frac{3^2}{b^2}=1;b=3$

To find $\text{a}$,

$c²=a²-b²=3²-3²=0$

Hence the equation is,

$\frac{{\left(x-1\right)}^2}{3²}+\frac{{\left(y-2\right)}^2}{3²}=1$

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