Chapter 10.4, Problem 72AYU

To determine

To find: Analyze the equation.

Answer to Problem 72AYU

The equation represents Ellipse.

Explanation of Solution

Given:

$x^2+36y^2-2x+288y+541=0$

Formula used:

Equation	Vertex	Focus	Major axis	Description
$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ $a>b>0 \text{and} b^2=a^2-c^2$	$\left(h\pm a,k\right)$	$\left(h\pm c,k\right)$	$x\text{-axis}$	Major axis is parallel to $x\text{-axis}$.

Calculation:

Given that $x^2+36y^2-2x+288y+541=0$.

$\left(x^2-2x\right)+36\left(y^2+8y\right)+541=0$

${\left(x-1\right)}^2+36{\left(y+4\right)}^2+541=1+576$

${\left(x-1\right)}^2+36{\left(y+4\right)}^2=36$

$\frac{{\left(x-1\right)}^2}{36}+{\left(y+4\right)}^2=1$

The equation represents an ellipse.

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

Center: $\left(h,k\right)=\left(1,-\text{4}\right)$.

Vertices: $V\left(h\pm a,k\right)=\left(1\pm 6,-4\right);\left(7,-4\right),\left(-5,-4\right)$.

$c^2=a^2-b^2$

$c^2={\text{ 6}}^2-{\text{ 1}}^2$; $c=\sqrt{35}$

Foci: $F\left(h\pm c,k\right)=F\left(1\pm \sqrt{35},-4\right)$.

The graph of the equation of $x^2+36y^2-2x+288y+541=0$ is plotted.

Using the given equation $x^2+36y^2-2x+288y+541=0$, we plot the graph using graphing tool.

The graph represents an ellipse, with vertex $\left(7,-\text{4}\right)$, $\left(-5,-\text{4}\right)$, the Major axis is parallel to $x\text{-axis}$.