Problem 40 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ 25 y^{2}+9 x^{2}-50 y-54 x=119 $$

Step-by-Step Solution

Answer

The equation is an ellipse: \( \frac{(y - 1)^2}{9} + \frac{(x - 3)^2}{25} = 1 \).

Step 1: Rearrange Terms

To convert the given equation into standard form, group the terms with variables on one side and the constant on the other. Start with the equation: \[ 25y^2 + 9x^2 - 50y - 54x = 119 \]Rearrange it so all the variable terms are on the left side and the constant on the right side remains the same:

Step 2: Complete the Square - Y Terms

Complete the square for the terms involving \(y\). For the expression \(25y^2 - 50y\), factor out the 25:\[ 25(y^2 - 2y) \] Take half of \(-2\), square it, to get \(1\). Add and subtract inside the bracket:\[ 25(y^2 - 2y + 1 - 1) = 25((y - 1)^2 - 1) = 25(y - 1)^2 - 25 \]

Step 3: Complete the Square - X Terms

Complete the square for the terms involving \(x\). For the expression \(9x^2 - 54x\), factor out the 9:\[ 9(x^2 - 6x) \] Take half of \(-6\), square it, to get \(9\). Add and subtract inside the bracket:\[ 9(x^2 - 6x + 9 - 9) = 9((x - 3)^2 - 9) = 9(x - 3)^2 - 81 \]

Step 4: Substitute and Simplify

Substitute the completed square expressions back into the equation:\[ 25(y - 1)^2 - 25 + 9(x - 3)^2 - 81 = 119 \]Combine like terms and move constants on the left to the right:\[ 25(y - 1)^2 + 9(x - 3)^2 = 225 \]

Step 5: Put in Standard Form of an Ellipse

Divide the entire equation by 225 to put it into standard form:\[ \frac{25(y - 1)^2}{225} + \frac{9(x - 3)^2}{225} = 1 \]Simplify:\[ \frac{(y - 1)^2}{9} + \frac{(x - 3)^2}{25} = 1 \]

Step 6: Identify the Conic Section

The equation \( \frac{(y - 1)^2}{9} + \frac{(x - 3)^2}{25} = 1 \) is in the standard form of an ellipse \( \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1 \) where \(a^2 = 9\) and \(b^2 = 25\). That's an ellipse because the denominators differ and both are positive.

Step 7: Graph the Ellipse

Plot the center of the ellipse at \((h, k) = (3, 1)\). The semi-major axis length \((b)\) is 5 (along the x-axis) and the semi-minor axis length \((a)\) is 3 (along the y-axis). Draw the ellipse using these dimensions around the center.

Key Concepts

Ellipse

An ellipse is a type of conic section that appears as an elongated circle. In simpler terms, it's a circle that has been stretched along one axis. This shape is defined by the equation: \[ \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1 \] Here, \(a^2\) and \(b^2\) determine the lengths of the axes of the ellipse, while \(h\) and \(k\) are the coordinates of its center.

• The larger denominator matches with the longer axis.
• An ellipse becomes circular if \(a = b\).

Understanding an ellipse is crucial for geometry because it shows up in orbits, optical applications, and more.

Standard Form

The standard form of an equation makes recognizing and analyzing geometric shapes much easier. For an ellipse, the standard form is: \[ \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1 \] This structure reveals important properties directly at a glance.

• \(h, k\) are the center of the ellipse.
• \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively.

Putting equations into standard form helps us discern the type of conic section (like ellipses) and its key attributes, without needing further computation.

Completing the Square

Completing the square is a method used to transition a quadratic equation into a perfect square trinomial. This is especially helpful when working with conic sections, like ellipses, where standard form is desired.For instance, converting \(25y^2 - 50y\) involved these steps:

• Factor out the coefficient from the squared term: \(25(y^2 - 2y)\)
• Identify the term to make a perfect square: Half of \(-2\) is \(-1\), which squared gives 1.
• Adjust within the brackets: \((y^2 - 2y + 1 - 1)\)
• Rewrite as a perfect square: \((y - 1)^2 - 1\)

This technique is crucial for rewriting quadratic terms, ensuring they fit a standard conic section format.

Graphing Conics

Graphing conic sections allows you to visually interpret how these equations correspond to shapes like ellipses, parabolas, circles, and hyperbolas. For ellipses, here's what to consider:

• Center: Located at \((h, k)\), from the transformed equation.
• Semi-major and semi-minor axes: Decide the stretch along each axis from the denominators \(a^2\) and \(b^2\).
• Orientation: The ellipse elongates along the axis with the greater denominator.

In our problem, the center of the ellipse is at \(3, 1\) with lengths of 5 and 3 along the x and y, respectively. Practicing graphing helps build intuition about the spatial properties of conic sections.