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 horizontal ellipse standard form:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{{\left(x-1\right)}^2}{36}+\frac{{\left(y-3\right)}^2}{16}=1$$
$$a^2=36$$
Take the square root of both sides of the equation:
$$a=6$$

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

In a horizontal ellipse, the major axis runs parallel to the x-axis and passes through the ellipse's vertices. Find the vertices by adding and subtracting $$a$$ from the x-coordinate $$\left(h\right)$$ of the center.

To find vertex_1, add $$a$$ to the x-coordinate $$\left(h\right)$$ of the center:
Vertex_1: $$\left(h+a,k\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$a=6$$
Vertex_1: $$\left(1+6,3\right)$$
Vertex_1: $$\left(7,3\right)$$

To find vertex_2, subtract $$a$$ from the x-coordinate ($$h$$) of the center:
Vertex_2: $$\left(h-a,k\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$a=6$$
Vertex_2: $$\left(1-6,3\right)$$
Vertex_2: $$\left(-5,3\right)$$

$$b$$ 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 horizontal ellipse standard form:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{{\left(x-1\right)}^2}{36}+\frac{{\left(y-3\right)}^2}{16}=1$$
$$b^2=16$$
Take the square root of both sides of the equation:
$$b=4$$
Because b represents a distance, it only has a positive value.

In a horizontal ellipse, the minor axis runs parallel to the y-axis and passes through the ellipse's co-vertices.
Find the co-vertices by adding and subtracting $$b$$ from the y-coordinate $$\left(k\right)$$ of the center.

To find co-vertex_1, add $$b$$ to the y coordinate $$\left(k\right)$$ of the center:
Co-vertex_1: $$\left(h,k+b\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$b=4$$
Co-vertex_1: $$\left(1,3+4\right)$$
Co-vertex_1: $$\left(1,7\right)$$

To find co-vertex_2, subtract $$b$$ from the y-coordinate $$\left(k\right)$$ of the center:
Co-vertex_2: $$\left(h,k-b\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$b=4$$
Co-vertex_2: $$\left(1,3-4\right)$$
Co-vertex_2: $$\left(1,-1\right)$$

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:
$$f=\sqrt{a^2-b^2}$$
$$a^2=36$$
$$b^2=16$$
Plug $$a^2$$ and $$b^2$$ into the formula and simplify:

$$f=\sqrt{36-16}$$

$$f=\sqrt{20}$$

$$f=4.472$$

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

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

To find focus_1, add $$f$$ to the x-coordinate $$\left(h\right)$$ of the center:
Focus_1: $$\left(h+f,k\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$f=4.472$$
Focus_1: $$\left(1+4.472,3\right)$$
Focus_1: $$\left(5.472,3\right)$$

To find focus_2, subtract $$f$$ from the x-coordinate $$\left(h\right)$$ of the center:
Focus_2: $$\left(h-f,k\right)$$
Center: $$\left(h,k\right)=\left(1,3\right)$$
$$h=1$$
$$k=3$$
$$f=4.472$$
Focus_2: $$\left(1-4.472,3\right)$$
Focus_2: $$\left(-3.472,3\right)$$

Use the formula for the area of an ellipse to find the ellipse's area:
$$\pi ·a·b$$
$$a=6$$
$$b=4$$
Plug $$a$$ and $$b$$ into the formula and simplify:

$$\pi ·6·4$$

$$\pi ·24$$

The area equals $$24\pi$$

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.

$$\frac{{\left(x-1\right)}^2}{36}+\frac{{\left(y-3\right)}^2}{16}=1$$

$$\frac{{\left(x-1\right)}^2}{36}+\frac{{\left(0-3\right)}^2}{16}=1$$

$$x_1=4.969$$

$$x_2=-2.969$$

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.

$$\frac{{\left(x-1\right)}^2}{36}+\frac{{\left(y-3\right)}^2}{16}=1$$

$$\frac{{\left(0-1\right)}^2}{36}+\frac{{\left(y-3\right)}^2}{16}=1$$

$$y_1=6.944$$

$$y_2=-0.944$$

To find the eccentricity use the formula:
$$\frac{\sqrt{a^2-b^2}}{a}$$
$$a^2=36$$
$$b^2=16$$
$$a=6$$
Plug $$a^2$$ , $$b^2$$ and $$a$$into the formula:

$$\frac{\sqrt{36-16}}{6}$$

$$\frac{\sqrt{20}}{6}$$

$$\frac{4.472}{6}$$

$$0.745$$

The eccentricity equals $$0.745$$