Chapter 10, Problem 8RE

In Problems 1-10, identify each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci: if it is a hyperbola, give its center, vertices, foci, and asymptotes.

$\text{4}{x}^2+\text{ 9}{y}^2-\text{ 16}x-\text{ 18}y\text{ = }11$

To determine

Each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

Answer to Problem 8RE

Ellipse

Center: $\left(2,1\right)$

Vertices: $\left(5,1\right),\left(\mathrm{-1},1\right)$

Foci: $\left(2\pm \surd 5,1\right)$.

Explanation of Solution

Given:

$4x²+9y²-16x-18y=11$

Formula used:

Center	Major axis	Foci	Vertices	Equation
$\left(h,k\right)$	Parallel to $x\text{-axis}$	$\left(h+c,k\right)$	$\left(h+a,0\right)$	$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ $a>b>0\text{ and }b²=a²-c²$
$\left(h,k\right)$	Parallel to $x\text{-axis}$	$\left(h-c,k\right)$	$\left(h-a,0\right)$

Calculation:

The equation $4x²+9y²-16x-18y=11$ can be rewritten as,

$4{\left(x-2\right)}^2+9{\left(y-1\right)}^2=11+16+9$

$\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{4}=1$

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

It represents an ellipse.

$a=3,b=2,h=2,k=1$

Center at $\left(h,k\right)=\left(2,\text{ 1}\right)$.

Vertices: $\left(h\pm a,k\right)=\left(2\pm 3,1\right)=\left(5,1\right),\left(-1,1\right)$.

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

$c=\sqrt{9-4}=\surd 5$

Foci: $\left(h\pm c,k\right)=\left(2\pm \surd 5,1\right)$.