Solution - Properties of ellipses

$$h$$ represents the x-offset from the origin.
$$k$$ represents the y-offset from the origin.
To find the values of $$h$$ and $$k$$, use the horizontal ellipse standard form:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$
$$h=0$$
$$k=0$$
Center: $$\left(0,0\right)$$

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

$$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$
$$a^2=\frac{1}{9}$$
Take the square root of both sides of the equation:
$$a=0.333$$

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: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$a=0.333$$
Vertex_1: $$\left(0+0.333,0\right)$$
Vertex_1: $$\left(0.333,0\right)$$

To find vertex_2, subtract $$a$$ from the x-coordinate ($$h$$) of the center:
Vertex_2: $$\left(h-a,k\right)$$
Center: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$a=0.333$$
Vertex_2: $$\left(0-0.333,0\right)$$
Vertex_2: $$\left(-0.333,0\right)$$

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

$$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$
$$b^2=\frac{1}{12}$$
Take the square root of both sides of the equation:
$$b=0.289$$
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: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$b=0.289$$
Co-vertex_1: $$\left(0,0+0.289\right)$$
Co-vertex_1: $$\left(0,0.289\right)$$

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: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$b=0.289$$
Co-vertex_2: $$\left(0,0-0.289\right)$$
Co-vertex_2: $$\left(0,-0.289\right)$$

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:
$$f=\sqrt{a^2-b^2}$$
$$a^2=\frac{1}{9}$$
$$b^2=\frac{1}{12}$$
Plug $$a^2$$ and $$b^2$$ into the formula and simplify:

$$f=\sqrt{\frac{1}{9}-\frac{1}{12}}$$

$$f=\sqrt{\frac{1}{36}}$$

$$f=0.167$$

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: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$f=0.167$$
Focus_1: $$\left(0+0.167,0\right)$$
Focus_1: $$\left(0.167,0\right)$$

To find focus_2, subtract $$f$$ from the x-coordinate $$\left(h\right)$$ of the center:
Focus_2: $$\left(h-f,k\right)$$
Center: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$f=0.167$$
Focus_2: $$\left(0-0.167,0\right)$$
Focus_2: $$\left(-0.167,0\right)$$

Use the formula for the area of an ellipse to find the ellipse's area:
$$\pi ·a·b$$
$$a=0.333$$
$$b=0.289$$
Plug $$a$$ and $$b$$ into the formula and simplify:

$$\pi ·0.333·0.289$$

$$\pi ·0.096$$

The area equals $$0.096\pi$$

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.

$$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$

$$\frac{x^2}{\frac{1}{9}}+\frac{0^2}{\frac{1}{12}}=1$$

$$x_1=\frac{1}{3}$$

$$x_2=-\frac{1}{3}$$

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.

$$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$

$$\frac{0^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{12}}=1$$

$$y_1=0.289$$

$$y_2=-0.289$$

To find the eccentricity use the formula:
$$\frac{\sqrt{a^2-b^2}}{a}$$
$$a^2=\frac{1}{9}$$
$$b^2=\frac{1}{12}$$
$$a=0.333$$
Plug $$a^2$$ , $$b^2$$ and $$a$$into the formula:

$$\frac{\sqrt{\frac{1}{9}-\frac{1}{12}}}{0.333}$$

$$\frac{\sqrt{\frac{1}{36}}}{0.333}$$

$$\frac{\frac{1}{6}}{0.333}$$

$$\frac{500}{999}$$

The eccentricity equals $$0.502$$