Problem 39 - Precalculus with Limits

Question

In Exercises 29-52, identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \((x+2)^2+\dfrac{(y+4)^2}{1/4}=1\)

Step-by-Step Solution

Answer

The conic is identified as an ellipse. The center is at (2,4), the semi-major axis a is 1, the semi-minor axis b is 1/2. The vertices are at points (1,4) and (3,4). The foci are at points (2+\( \frac {\sqrt 3}{2}\), 4) and (2-\( \frac {\sqrt 3}{2}\), 4). The eccentricity is \( \frac {\sqrt 3}{2}\).

Step 1: Identify the Conic

The given equation \((x+2)^2+\dfrac{(y+4)^2}{1/4}=1\) is not in the form of a circle but of an ellipse since the denominators of \(x\) and \(y\) squares are not equal. So, it's an ellipse.

Step 2: Find the Center

The center of the ellipse is given by \(-h, -k\) , which in this case is \(-2, -4\). So, the center is at (2,4).

Step 3: Determine the Semi-Axes

For the standard equation, a^2 is under the \(x\) term and b^2 is under the \(y\) term. Hence, \(a^2=1\) and \(b^2=1/4\). So, the semi-major axis \(a=1\) and the semi-minor axis \(b=1/2\). The vertices are endpoints of major axis, that are at a distance a from center. Therefore, the vertices are at (2±1,4) which are at points (1,4) and (3,4).

Step 4: Calculate the Foci and Eccentricity

The foci for an ellipse lie on the major axis at a distance of \(c=\sqrt{a^2-b^2}\) from the center. In this case, it becomes \(\sqrt{1-(1/2)^2}=\sqrt{3/4}\), which simplifies to \(c = \dfrac{\sqrt{3}}{2}\). Hence, the foci are at points \((2±\dfrac{\sqrt{3}}{2}, 4)\). The eccentricity (e) of an ellipse is given by \(e=c/a\), substituting the given values we have \(e=\dfrac{\sqrt{3}}{2}/1=\dfrac{\sqrt{3}}{2}\).

Step 5: Sketch the Graph

Now, plot the center, vertices and foci on a graph. Draw an ellipse through these points. The center is at (2,4), the vertices are at (1,4) and (3,4), and the foci are at \((2±\dfrac{\sqrt{3}}{2}, 4)\).

Key Concepts

Ellipse Equation

An ellipse is a set of points in a plane, the sum of whose distances from two fixed points, known as foci, is constant. Understanding its equation is crucial to studying its properties. The standard form of an ellipse equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \( (h,k) \) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. If \( a^2 > b^2 \), the ellipse stretches horizontally, otherwise vertically.

Applying this to the given exercise, \( (x+2)^2 + \frac{(y+4)^2}{1/4} = 1 \) is identified as an ellipse equation because the denominators of the squared terms \( x \) and \( y \) are distinct, indicating varying axis lengths. Therefore, this equation represents an ellipse with a horizontal orientation due to \( a^2 > b^2 \).

Vertices of Ellipse

Vertices are pivotal points that define the 'widest' sections of an ellipse. They can be found by examining the ellipse equation and determining the semi-major axis length \(a\). The vertices are positioned at a distance \( a \) from the center along the major axis.

In our exercise, with the center being at (-2, -4) and \( a = 1\), the vertices can be pinpointed by moving 1 unit horizontally from the center, since our \( a^2 \) is under the \( x \) term. This results in the vertices being at (1, 4) and (3, 4), which are exactly \( a \) units to the left and right from the center of the ellipse.

Foci of Ellipse

The foci of an ellipse are two points located along the major axis, equidistant from the center. The distance from the center to each focus, denoted as \(c\), is given by the equation \(c = \sqrt{a^2 - b^2}\). The ellipse's defining property is that the sum of the distances from any point on the ellipse to the foci is constant.

For the ellipse in our exercise, \(c\) is calculated to be \(\frac{\sqrt{3}}{2}\), signifying that the foci are at points \( (2 \pm \frac{\sqrt{3}}{2}, 4) \). These points are \(\frac{\sqrt{3}}{2}\) units left and right from the center on the horizontal axis.

Eccentricity of Ellipse

Eccentricity is a measure that describes how much an ellipse deviates from being a circle. It ranges from 0 for a perfect circle to almost 1 for an elongated ellipse. Eccentricity, denoted as \(e\), is defined by the ratio \( e = c/a \), where \(c\) is the distance from the center to a focus and \(a\) is the semi-major axis.

In the exercise, the eccentricity \(e\) turns out to be \(\frac{\sqrt{3}}{2}\) upon substituting the values of \(c\) and \(a\). This tells us the ellipse is slightly elongated since the value is closer to 1 than it is to 0, but not extremely so, since it is still less than 1.