Chapter 10, Problem 10RE

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.

$\text{9}{x}^2+\text{ 4}{y}^2-\text{ 18}x\text{ + 8y = }23$

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 10RE

Ellipse

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

Vertices: $\left(1,\mathrm{-4}\right),\left(1,2\right)$

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

Explanation of Solution

Given:

$9x²+4y²-18x+8y=23$

Formula used:

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

Calculation:

The equation $9x²+4y²-18x+8y=23$ can be rewritten as,

$9{\left(x-1\right)}^2+4{\left(y+1\right)}^2=23+4+9$

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

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

It represents an ellipse.

$a=3,b=2,h=1,k=\mathrm{-1}$

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

Vertices: $\left(h,k\pm a\right)=\left(1,-1\pm 3\right)=\left(1,-4\right),\left(1,2\right)$.

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

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

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