Problem 37 - 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. \(\dfrac{x^2}{4/9}+\dfrac{(y+1)^2}{4/9}=1\)

Step-by-Step Solution

Answer

The given conic section is a circle centered at (0, -1) with a radius of \(2/3\) units. The vertices are located at (-2/3, -1) and (2/3,-1), the foci are at the center due to the circle being also an ellipse with eccentricity of 0.

Step 1: Identify the Conic

The general equation for an ellipse when the center is not at the origin is \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\). Here, the given equation fits this format, so we can already identify the conic section as an ellipse.

Step 2: Identify Center, Vertices, and Semi-Major Axis (a)

Here both \(x^2\) and \(y^2\) are divided by \(4/9\). This tells us that a=b=2/3, since a and b are the square roots of \(4/9\). We know that the center of the ellipse is at (h, k). Looking at the equation we can see the center is (0, -1), since there is no shift on the x-axis, implied by x alone, and y was shifted down by 1, indicated by \(y+1\). Vertices are at a distance 'a' from the center along the major axis. Here because a=b, the ellipse is a circle and there are no major or minor axes. The vertices are at the end points of the diameter of the circle which are at (-2/3, -1) and (2/3, -1)

Step 3: Identify Radius, Foci and Eccentricity (e)

Since our ellipse is a circle the foci are at the center (0, -1). The radius of the circle is equal to 'a' or 'b' which here is 2/3. For an ellipse, the eccentricity, e, can be found with the formula e = \sqrt{1 - (b^2 / a^2)}. Since here a = b, then b^2 / a^2 = 1, making e = 0.

Step 4: Sketch the Ellipse (which is a Circle in this case)

Finally, with the above information, we can sketch the ellipse. It will be a circle, centered at (0,-1), with radius of 2/3 units.

Key Concepts

Circle

A circle is a unique type of ellipse where the eccentricity is zero, meaning it is perfectly round with no elongation. In mathematical terms, any equation of a circle can be expressed in the form \[(x-h)^2 + (y-k)^2 = r^2\] where

• \((h, k)\) is the center of the circle.
• \(r\) represents the radius.

In our exercise, although the equation initially looks like an ellipse, both the \(x\) and \(y\) terms are divided by the same constant \(\left(\frac{4}{9}\right)\). This signals that the ellipse is actually a circle. The circle is centered at \((0, -1)\), with a radius of \(\frac{2}{3}\). Circles are considered as the most symmetric shapes in Euclidean geometry.

Eccentricity

The concept of eccentricity measures how much a conic section deviates from being circular. For ellipses, it is a number between 0 and 1, where 0 is a perfect circle, and values closer to 1 are more stretched. The formula for eccentricity \(e\) of an ellipse is\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]where \(a\) and \(b\) are the semi-major and semi-minor axes respectively.

• In our exercise, since both semi-major and semi-minor axes are equal, \(a = b\), making \(b^2/a^2 = 1\).
• This results in \(e = 0\), confirming that our conic is a circle which has no eccentricity.

Center of a Conic

The center of a conic is a crucial point from which symmetry and other important properties stem. In a standard equation \[(x-h)^2/a^2 + (y-k)^2/b^2 = 1\]

• The center is \((h, k)\).

In our given equation, there is no shift in the \(x\)-direction and a shift of \(-1\) in the \(y\)-direction, leading to a center of \((0, -1)\). The center acts like the balance point for the conic.

Vertices of a Conic

Vertices represent the points where the conic reaches its maximum length along the axes. For circles:

• The vertices lie at the ends of its diameter.

In our case, this circle's vertices are calculated at a distance of the radius along the axis of symmetry. Since the center is \((0, -1)\) with a radius \(\frac{2}{3}\), the vertices are located at

• \((-\frac{2}{3}, -1)\) and \((\frac{2}{3}, -1)\)
.

These points give the furthest reach of the circle along the axis.

Foci of a Conic

Foci are special points located along the major axis of an ellipse. They play a key role in defining the shape of ellipses. However, for circles:

• The foci coincide with the center, due to their zero eccentricity.

This means every point on the circumference is equidistant from the center. As a result, in a perfectly symmetric circle like ours, the foci do not provide additional distance information. Here, the foci and the center both share the location \((0, -1)\). This indicates that the circular shape means uniform distance from the center to any point on the circle.