Question

Find an equation of the following ellipses, assuming the center is at the origin. Sketch a graph labeling the vertices and foci. An ellipse with vertices \((0,\pm 10),\) passing through the point \((\sqrt{3} / 2,5)\)

Short answer

Given an ellipse whose vertices are at points (0, ±10) and which passes through the point (sqrt(3)/2, 5), we first determined its equation as (3x^2)/4 + y^2/100 = 1. The semi-major axis is 10, and the semi-minor axis is sqrt(4/3). The foci are at points (0, ± sqrt(296)/3). When sketching the ellipse, make sure to mark the vertices, foci, and ensure the ellipse passes through the given point (sqrt(3)/2, 5).

Step-by-step solution

Identify the Given Information

The center of the ellipse is at the origin, and its vertices are at \((0, \pm 10)\). The vertices lie on the vertical axis, so the ellipse is elongated vertically. From the vertices, we can determine the length of the semi-major axis as 10, thus \(a = 10\). The general equation of an ellipse elongated vertically centered at the origin is given by: $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ Substitute the given value of \(a\): $$\frac{x^2}{b^2} + \frac{y^2}{100} = 1$$

Use the Point Given to Find the Value of \(b^2\)

We know that the ellipse passes through the point \((\sqrt{3} / 2, 5)\). Therefore, this point must satisfy the general equation of the ellipse found in the previous step. Substitute the point \((\sqrt{3} / 2, 5)\) in this equation: $$\frac{(\frac{\sqrt{3}}{2})^2}{b^2} + \frac{5^2}{100} = 1$$ Now, solve for \(b^2\): $$\frac{3}{4b^2} + \frac{25}{100} = 1$$ $$\frac{3}{4b^2} = \frac{75}{100}$$ By isolating, we find the value of \(b^2\): $$b^2 = \frac{4}{3}$$

Find the Equation of the Ellipse

Now, substitute the value of \(b^2\) back into the general equation of the ellipse: $$\frac{x^2}{\frac{4}{3}} + \frac{y^2}{100} = 1$$ Hence, the equation of the ellipse is: $$\frac{3x^2}{4} + \frac{y^2}{100} = 1$$

Determine the Value of the Semi-Minor Axis

Now that we have found the equation of the ellipse, we can determine the value of the semi-minor axis, \(b\), by taking the square root of \(b^2\): $$b = \sqrt{\frac{4}{3}}$$

Identify the Foci Points

Since the ellipse is elongated vertically, we will use the following formula to find the distance from the center to the foci, denoted as \(c\): $$c = \sqrt{a^2 - b^2}$$ Substitute the values of \(a\) and \(b^2\): $$c = \sqrt{100 - \frac{4}{3}}$$ Evaluating the expression results in: $$c = \sqrt{\frac{296}{3}}$$ The foci will be at points \((0, \pm c)\) or \((0, \pm \frac{\sqrt{296}}{3})\).

Sketch the Graph and Label the Vertices and Foci

To sketch the graph with labeled vertices and foci, follow these steps: 1. Plot the origin (0,0) as the center of the ellipse. 2. Draw the vertical major axis of length 2a, which is equal to 20 units. 3. Draw the horizontal minor axis of length 2b, which is equal to 2 * \(\sqrt{\frac{4}{3}}\). 4. Mark the vertices (0, ±10). 5. Mark the foci (0, ± \(\frac{\sqrt{296}}{3}\)). 6. Sketch the ellipse, ensuring it passes through the given point \((\sqrt{3} / 2, 5)\). With these steps, the graph is now complete with the ellipse's equation, vertices, and foci labeled and shown.

Key concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These curves include circles, ellipses, parabolas, and hyperbolas. Each shape has unique properties related to its geometric structure.
An ellipse, a type of conic section, appears as an elongated circle and is defined mathematically by its sum of distances from two fixed points, called foci, being constant. This intriguing aspect makes ellipses valuable in astronomy and physics, such as orbit paths of planets.

Semi-Major Axis

The semi-major axis is one of the most important aspects of an ellipse. It represents the longest radius of the ellipse, extending from its center to the furthest point on its path. For the given ellipse problem, the semi-major axis is 10 units long, aligning vertically.
This axis is crucial in determining the overall dimensions and shape of the ellipse. In any ellipse, the semi-major axis is denoted by the letter \(a\), and it sets how stretched the ellipse appears.

Semi-Minor Axis

The semi-minor axis is the shorter radius of an ellipse and extends from the center to the edge, in the direction perpendicular to the semi-major axis. It helps in defining the ellipse's width.
In our problem, the semi-minor axis is calculated from the equation derived by substituting given points, which results in \(b^2 = \frac{4}{3}\). The actual length \(b\) becomes \(\sqrt{\frac{4}{3}}\), defining how narrow the ellipse is compared to its length.

Foci

Foci are two fixed points on the interior of an ellipse. The sum of the distances from any point on the ellipse to the foci is constant, providing a vital mathematical property for defining ellipses.
For the given case, the foci are found using the formula \(c = \sqrt{a^2 - b^2}\), where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. This calculation, \(c = \sqrt{\frac{296}{3}}\), gives us the precise location of the foci, marked along the major axis.

Equation of an Ellipse

The equation of an ellipse provides a concise way to describe its geometric structure mathematically. For ellipses centered at the origin, the standard form is:
\(rac{x^2}{b^2} + rac{y^2}{a^2} = 1\) for vertical ellipses.
By substituting values into this equation from our problem, we attain \(\frac{3x^2}{4} + \frac{y^2}{100} = 1\). This equation is essential for plotting the ellipse accurately and analyzing its properties, such as intercepts and axes.