Problem 35 - Precalculus with Calculus Previews

Question

Find an equation of the ellipse that satisfies the given conditions. Vertices (±4,1) , passing through \((2 \sqrt{3}, 2)\)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{16} + \frac{(y-1)^2}{4} = 1\).

Step 1: Identify the Center and Axes Directions

The given vertices are \((\pm 4,1)\), which implies that the ellipse is centered at \((0,1)\). The vertices are horizontally aligned, indicating that the major axis is horizontal.

Step 2: Determine the Length of the Major Axis

The distance between the vertices \((4, 1)\) and \((-4, 1)\) is 8. Thus, the length of the major axis \(2a\) is 8, leading to \(a = 4\).

Step 3: Determine the General Equation of the Ellipse

The standard form of the ellipse's equation when the major axis is horizontal is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Substituting the center \((h, k) = (0,1)\) and \(a = 4\), we have \(\frac{x^2}{16} + \frac{(y - 1)^2}{b^2} = 1\).

Step 4: Use the Point (2\sqrt{3}, 2) to Find b

Substitute \(x = 2\sqrt{3}\) and \(y = 2\) into the ellipse equation: \(\frac{(2\sqrt{3})^2}{16} + \frac{(2-1)^2}{b^2} = 1\). This simplifies to \(\frac{12}{16} + \frac{1}{b^2} = 1\).

Step 5: Solve the Equation for b^2

Solve the equation \(\frac{3}{4} + \frac{1}{b^2} = 1\) by first subtracting \(\frac{3}{4}\) from both sides to get \(\frac{1}{b^2} = \frac{1}{4}\). Thus, \(b^2 = 4\).

Step 6: Write the Final Equation

Plugging \(b^2 = 4\) back into the ellipse equation gives the final equation: \(\frac{x^2}{16} + \frac{(y - 1)^2}{4} = 1\).

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. Depending on the angle and position of the intersection, different types of curves are formed, such as circles, ellipses, parabolas, and hyperbolas. These are fundamental shapes in geometry and their properties are widely utilized in various fields like physics, engineering, and astronomy.

An ellipse is one of these intriguing conic sections. It is formed when the plane cuts through the cone at an angle that results in an oval shape, rather than a sharp corner like a parabola or two opposite arcs as seen in hyperbolas. Ellipses can be seen in orbits of planets and satellites, making them especially significant in celestial mechanics.

The general equation for an ellipse centered at the origin is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. This equation highlights how each conic shape has a unique equation that describes its set of points.

Vertices of Ellipse

The vertices of an ellipse are the endpoints of the major axis, which is the longest diameter of the ellipse. These points are crucial in defining the shape and size of the ellipse.

For an ellipse with its center at \((h, k)\) and major axis parallel to the x-axis, the vertices can be found at \((h\pm a, k)\). If the major axis is vertical, the vertices shift to \((h, k\pm a)\). This distinction is important as it determines the orientation and spread of the ellipse.

In the provided exercise, the vertices are located at \((\pm 4, 1)\). This indicates the center is at \((0,1)\) and the major axis is horizontal with a distance of 8 between the vertices, resulting in a semi-major axis \(a = 4\). Knowing the vertices helps in finding the equation of the ellipse, as they determine the value of \(a\).

Major and Minor Axes

An ellipse has two primary axes that define its shape: the major axis and the minor axis. The major axis is the longest line dividing the ellipse into two equal halves, while the minor axis is the shortest. These axes are perpendicular and intersect at the center of the ellipse.

The length of the major axis is \(2a\) and that of the minor axis is \(2b\), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. The orientations can vary:

• Horizontal Major Axis: Longer in the x-direction, with vertices at \((h\pm a, k)\).
• Vertical Major Axis: Longer in the y-direction, vertices at \((h, k\pm a)\).

The properties of the axes are essential to understand as they affect the equation of the ellipse.

In our example, the major axis is horizontal, determined by vertices at \((\pm 4, 1)\). Solving the problem involves using the lengths \(a=4\) and \(b\) derived from the equation, ultimately leading to knowing \(b^2 = 4\), which aligns with the minor axis length.