Problem 17 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 25 x^{2}+9 y^{2}-100 x+18 y-116=0 $$

Step-by-Step Solution

Answer

The center is (2,-1), vertices (2,-6) and (2,4), foci (2,-5) and (2,3), minor axis endpoints (-1,-1) and (5,-1), and eccentricity \(e\) is 0.8.

Step 1: Rewrite the Equation in Standard Form

Start by rewriting the given equation in its standard ellipse form. The given equation: \(25x^2 + 9y^2 - 100x + 18y - 116 = 0\).Group the \(x\) and \(y\) terms: \( (25x^2 - 100x) + (9y^2 + 18y) = 116 \).Complete the square for each group.

Step 2: Complete the Square for x Terms

For the \(x\) terms, \(25x^2 - 100x\), factor out the 25:\[ 25(x^2 - 4x) \].Complete the square: Take half of the linear term (4), square it (4), and add/subtract inside the bracket:\[ 25(x^2 - 4x + 4 - 4) = 25((x-2)^2 - 4) \].This becomes \(25(x-2)^2 - 100\).

Step 3: Complete the Square for y Terms

For the \(y\) terms, \(9y^2 + 18y\), factor out the 9:\[ 9(y^2 + 2y) \].Complete the square:Take half of the linear term (2), square it (1), and add/subtract inside the bracket:\[ 9(y^2 + 2y + 1 - 1) = 9((y+1)^2 - 1) \].This becomes \(9(y+1)^2 - 9\).

Step 4: Rewrite the Completed Squares and Simplify

Substitute the completed squares into the original equation:\[ 25(x-2)^2 - 100 + 9(y+1)^2 - 9 = 116 \]. Simplify the equation:\[ 25(x-2)^2 + 9(y+1)^2 = 225 \].Divide through by 225 to get the equation in standard form:\[ \frac{(x-2)^2}{9} + \frac{(y+1)^2}{25} = 1 \].

Step 5: Identify the Center and Axes Lengths

The standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) gives us the center at \((h, k)\), which is \((2, -1)\) here.From \(\frac{(x-2)^2}{9}\) and \(\frac{(y+1)^2}{25}\), we recognize that \(a^2 = 9\) and \(b^2 = 25\). Hence, \(a = 3\) and \(b = 5\).Since \(b > a\), this means the major axis is vertical.

Step 6: Find Vertices and Endpoints of the Minor Axis

The vertices are along the major axis at \( (h, k \pm b) \): \((2, -1 \pm 5)\) or \((2, -6)\) and \((2, 4)\).Endpoints of the minor axis are along \(x\) direction at \( (h \pm a, k) \): \((2 \pm 3, -1)\) or \((-1, -1)\) and \((5, -1)\).

Step 7: Find the Foci

Compute the eccentricity \( c \) using \( c^2 = b^2 - a^2 \):\( c^2 = 25 - 9 = 16 \), hence \( c = 4 \).The foci \((h, k \pm c)\) are \((2, -1 \pm 4)\): \((2, -5)\) and \((2, 3)\).

Step 8: Calculate the Eccentricity

Eccentricity \(e\) is given by \(e = \frac{c}{b}\):\(e = \frac{4}{5}\).This shows how "elongated" the ellipse is.

Key Concepts

Center of Ellipse

An ellipse is a shape that resembles an elongated circle, and its equation can describe positions on a plane. The center of the ellipse is a crucial point as it determines the position of the ellipse on the coordinate plane. In standard form, the equation of the ellipse is \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where

• \((h, k)\) is the center of the ellipse,
• \(a\) is the semi-major axis length, and
• \(b\) is the semi-minor axis length.

In the exercise, the standard form is determined to be \(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{25} = 1\), which instantly tells us that the center is at

• \((h, k) = (2, -1)\).

This point is critical, especially when graphing, as it helps to position the base of the axes of the ellipse.

Foci of Ellipse

Foci are special points on the major axis of an ellipse. These points are integral to the definition of an ellipse. The sum of the distances from any point on the ellipse to each focus is constant. In simpler terms, the placement of foci affects the shape of the ellipse. To find the foci, calculate \[c^2 = b^2 - a^2\]. For our problem, we found:

• \(b^2 = 25\) and \(a^2 = 9\),
• hence \(c^2 = 25 - 9 = 16\), resulting in \(c = 4\).

With the center at \((2, -1)\), the foci are located along the major axis, yielding

• \((2, -1 \pm 4)\), which gives \((2, -5)\) and \((2, 3)\).

These points are critical because they define the elliptical shape's longest span.

Vertices of Ellipse

Vertices are points where the ellipse is the widest; they sit on the ends of the major axis of the ellipse. For the vertices, we use the center and semi-major axis length calculated by

• \((h, k \pm b)\).
• In our case, \(b = 5\), placing vertices at \((2, -1 \pm 5)\).
• This simplifies to \((2, -6)\) and \((2, 4)\).

These vertices are important for sketching the ellipse as they mark the points where the ellipse reaches its maximum extension.

Eccentricity of Ellipse

Eccentricity measures how much the ellipse devotes from being a circle. It is defined as \[e = \frac{c}{b}\], where

• \(c\) is the focal distance, and
• \(b\) is the semi-major axis length.

In the context of the exercise,

• \(c = 4\) and \(b = 5\),
• so \(e = \frac{4}{5} = 0.8\).

This indicates that the ellipse is not very elongated but also not very close to a circle. Eccentricity values range from 0 (a perfect circle) to less than 1 (an ellipse that devotes from circular shape), making it a valuable tool in identifying the overall shape of the ellipse.