Solution - Properties of ellipses

$$a$$ represents the longer radius of the ellipse, which is equal to half of the major axis.
This is called the semi-major axis.
To find the value of $$a$$, use the vertical ellipse standard form:
$$\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$$

$$\frac{{\left(x-5\right)}^2}{9}+\frac{{\left(y+3\right)}^2}{16}=1$$
$$a^2=16$$
Take the square root of both sides of the equation:
$$a=4$$

Because $$a$$ represents a distance, it only has a positive value.

In a vertical ellipse, the major axis runs parallel to the y-axis and passes through the ellipse's vertices. Find the vertices by adding and subtracting $$a$$ from the y-coordinate ($$k$$) of the center.

To find vertex_1, add $$a$$ to the y-coordinate ($$k$$) of the center:
Vertex_1: $$\left(h,k+a\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$a=4$$
Vertex_1: $$\left(5,-3+4\right)$$
Vertex_1: $$\left(5,1\right)$$

To find vertex_2, subtract $$a$$ from the y-coordinate ($$k$$) of the center:
Vertex_2: $$\left(h,k-a\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$a=4$$
Vertex_2: $$\left(5,-3-4\right)$$
Vertex_2: $$\left(5,-7\right)$$

$$b$$ represents the shorter radius of the ellipse, which is equal to half of the minor axis. This is called the semi-minor axis.
To find the value of $$b$$, use the vertical ellipse standard form:
$$\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$$

$$\frac{{\left(x-5\right)}^2}{9}+\frac{{\left(y+3\right)}^2}{16}=1$$
$$b^2=9$$
Take the square root of both sides of the equation:
$$b=3$$
Because b represents a distance, it only has a positive value.

In a vertical ellipse, the minor axis runs parallel to the x-axis and passes through the ellipse's co-vertices.
Find the co-vertices by adding and subtracting $$b$$ from the x-coordinate ($$h$$) of the center.

To find co-vertex_1, add $$b$$ to the x-coordinate ($$h$$) of the center:
Co-vertex_1: $$\left(h+b,k\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$b=3$$
Co-vertex_1: $$\left(5+3,-3\right)$$
Co-vertex_1: $$\left(8,-3\right)$$

To find co-vertex_2, subtract $$b$$ from the x-coordinate ($$h$$) of the center:
Co-vertex_2: $$\left(h-b,k\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$b=3$$
Co-vertex_2: $$\left(5-3,-3\right)$$
Co-vertex_2: $$\left(2,-3\right)$$

Focal length is the distance from the ellipse's center to each focal point and is usually represented by $$f$$.

To find $$f$$, use the formula:
$$f=\sqrt{a^2-b^2}$$
$$a^2=16$$
$$b^2=9$$
Plug $$a^2$$ and $$b^2$$ into the formula and simplify:

$$f=\sqrt{16-9}$$

$$f=\sqrt{7}$$

$$f=2.646$$

Because $$f$$ represents a distance, it only has a positive value.

In a vertical ellipse, the major axis runs parallel to the y-axis and through the foci.
Find the foci by adding and subtracting $$f$$ from the y-coordinate $$\left(k\right)$$ of the center.

To find focus_1, add $$f$$ to the y-coordinate $$\left(k\right)$$ of the center:
Focus_1: $$\left(h,k+f\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$f=2.646$$
Focus_1: $$\left(5,-3+2.646\right)$$
Focus_1: $$\left(5,-0.354\right)$$

To find focus_2, subtract $$f$$ from the y-coordinate $$\left(k\right)$$ of the center:
Focus_2: $$\left(h,k-f\right)$$
Center: $$\left(h,k\right)=\left(5,-3\right)$$
$$h=5$$
$$k=-3$$
$$f=2.646$$
Focus_2: $$\left(5,-3-2.646\right)$$
Focus_2: $$\left(5,-5.646\right)$$

Use the formula for the area of an ellipse to find the ellipse's area:
$$\pi ·a·b$$
$$a=4$$
$$b=3$$
Plug $$a$$ and $$b$$ into the formula and simplify:

$$\pi ·4·3$$

$$\pi ·12$$

The area equals $$12\pi$$

To find the x-intercept(s), plug $$0$$ in for $$y$$ in the ellipse's standard equation and solve the resulting quadratic equation for $$x$$.
Click here for a step-by-step explanation of the quadratic equation.

$$\frac{{\left(x-5\right)}^2}{9}+\frac{{\left(y+3\right)}^2}{16}=1$$

$$\frac{{\left(x-5\right)}^2}{9}+\frac{{\left(0+3\right)}^2}{16}=1$$

$$x_1=6.984$$

$$x_2=3.016$$

To find the y-intercept(s), plug $$0$$ in for $$x$$ in the ellipse's standard equation and solve the resulting quadratic equation for $$y$$.
Click here for a step-by-step explanation of the quadratic equation.

$$\frac{{\left(x-5\right)}^2}{9}+\frac{{\left(y+3\right)}^2}{16}=1$$

$$\frac{{\left(0-5\right)}^2}{9}+\frac{{\left(y+3\right)}^2}{16}=1$$

$$\sqrt{-}\frac{16}{9}$$

To find the eccentricity use the formula:
$$\frac{\sqrt{a^2-b^2}}{a}$$
$$a^2=16$$
$$b^2=9$$
$$a=4$$
Plug $$a^2$$ , $$b^2$$ and $$a$$into the formula:

$$\frac{\sqrt{16-9}}{4}$$

$$\frac{\sqrt{7}}{4}$$

$$\frac{2.646}{4}$$

$$0.661$$

The eccentricity equals $$0.662$$