Problem 27 - Precalculus Mathematics for Calculus

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$4 x^{2}+25 y^{2}-24 x+250 y+561=0$$

Step-by-Step Solution

Answer

The equation represents an ellipse centered at (3, -5) with vertices at (-2, -5) & (8, -5), minor axis vertices (3, -3) & (3, -7), and foci at \((3 \pm \sqrt{21}, -5)\).

Step 1: Rearrange and Group Terms

Start by moving the constant term to the other side of the equation. Rearrange the terms to group the x terms and the y terms together: \[4x^2 - 24x + 25y^2 + 250y = -561\]

Step 2: Complete the Square for x Terms

For the x terms \(4x^2 - 24x\), factor out 4 first to get \(4(x^2 - 6x)\). Complete the square inside the parenthesis: \[x^2 - 6x \rightarrow (x - 3)^2 - 9\]. Thus, we have \[4((x - 3)^2 - 9) = 4(x - 3)^2 - 36\].

Step 3: Complete the Square for y Terms

For the y terms \(25y^2 + 250y\), factor out 25 to get \(25(y^2 + 10y)\). Complete the square inside the parenthesis: \[y^2 + 10y \rightarrow (y + 5)^2 - 25\]. Thus, we have \[25((y + 5)^2 - 25) = 25(y + 5)^2 - 625\].

Step 4: Simplify the Equation

Substitute the completed squares back into the original equation: \[4(x - 3)^2 - 36 + 25(y + 5)^2 - 625 = -561\]. Simplify the equation: \[4(x - 3)^2 + 25(y + 5)^2 = 100\].

Step 5: Identify the Type of Conic

The equation \[4(x - 3)^2 + 25(y + 5)^2 = 100\] resembles the standard form of an ellipse \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]. It is an ellipse because both squared terms are positive, and the coefficients are of different magnitudes.

Step 6: Rewrite the Equation in Standard Form

Divide the entire equation by 100 to match the standard form of an ellipse: \[\frac{(x - 3)^2}{25} + \frac{(y + 5)^2}{4} = 1\]. Now, the ellipse is in the standard form.

Step 7: Find the Properties of the Ellipse

From the equation \[\frac{(x-3)^2}{5^2} + \frac{(y+5)^2}{2^2} = 1\], determine that the center is \((3, -5)\). The lengths of the axes are: \(a = 5\) and \(b = 2\). The vertices (major axis) are at \((3 \pm 5, -5)\) or \((-2, -5)\) and \((8, -5)\). The minor axis vertices are \((3, -5 \pm 2)\) or \((3, -3)\) and \((3, -7)\). The foci are \((3 \pm \sqrt{21}, -5)\), calculated as \(\pm \sqrt{a^2 - b^2} = \pm \sqrt{25 - 4} = \pm \sqrt{21}\).

Key Concepts

Ellipse

An ellipse is a curved shape that resembles a squashed circle. It is one of the four types of conic sections, the others being hyperbolas, parabolas, and circles. Ellipses can be described and analyzed using their standard form equation. A unique property of ellipses is that the sum of the distances from any point on the ellipse to two fixed points, known as foci, is constant.

• The center of an ellipse is the midpoint between its vertices.
• Major and minor axes are the longest and shortest diameters of the ellipse, respectively.
• The distance between the center and a focus is determined by the relation \(\sqrt{a^2 - b^2}\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes.

When identifying an ellipse from an equation after completing the square, it's crucial that both \(x^2\) and \(y^2\) terms are positive and different in magnitude.

Completing the Square

Completing the square is a method used to solve quadratic equations and to rewrite equations in a standard form such as that of an ellipse. This technique is incredibly useful when working with conic sections because it makes the process of identifying conic shapes much easier.
To complete the square:

• Group terms involving \(x\) and \(y\) separately.
• For a binomial expression like \(x^2 - 6x\), take half of the coefficient of \(x\), square it, and add inside the bracket: \( (x - 3)^2 - 9 \).
• Apply the same for \(y^2 + 10y\), resulting in \( (y + 5)^2 - 25 \).

Completing the square transforms equations into a more recognizable format, revealing whether the graph represents an ellipse, parabola, hyperbola, or another conic section. This step is critical for ensuring terms can be easily refactored into standard equations.

Equation of an Ellipse

The standard form of an ellipse's equation is crucial to identifying its key properties. For an ellipse centered at \( (h, k) \), its equation can be expressed as:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

• \((h, k)\) is the center of the ellipse.
• \((a)\) is the semi-major axis; \((b)\) is the semi-minor axis.
• If the squared term in the denominator of \((x-h)^2\) is larger, the ellipse is stretched horizontally. If larger in \((y-k)^2\), the stretch is vertical.

Recognizing this equation allows for identifying critical features:

• The vertices, where the ellipse intersects its axes.
• The foci, positioned along the major axis distance \( \sqrt{a^2-b^2} \) from the center.

This form is derived by completing the square for both x and y terms, ensuring the constant on the right side is 1 after dividing all terms in the equation by the result of completing the square.