Problem 56 - College Algebra

Question

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$4 x^{2}+25 y^{2}-24 x+100 y+36=0$$

Step-by-Step Solution

Answer

After completing the square in the equation, we get \(4(x-3)^2 + 25(y+2)^2 = 116\), the standard form becomes \((x-3)^2/29 + (y+2)^2/4.64 = 1\). After graphing the ellipse with center at (3, -2), a= \(\sqrt{29}\) and b=\(\sqrt{4.64}\), we then find the location of the foci to be at \(\sqrt{29-4.64}\) distance from the center.

Step 1: Reordering the Equation

First, rearrange the equation to place the \(x\)s and \(y\)s together: \(4x^{2}-24x+25y^{2}+100y=-36\)

Step 2: Completing the square for x and y

Now we complete the square for \(x\) and \(y\). This means making it in the form \(a(x-h)^{2}\) and \(b(y-k)^{2}\). For \(x\), the square is completed by taking half of the coefficient of \(x\), squaring it and adding it to the equation. Here, the coefficient of \(x\) is \(-24/4=-6\). When squared, it gives \(36\). The equation is now \(4(x^2 - 6x + 9) + 25y^{2}+100y=-36 + 4*9\). Doing the same operation for \(y\), we divide 100 by 50 giving \(2\) and square it to give \(4\). The equation becomes \(4(x-3)^{2}+25(y^2+4y+4)=0+16+100\).

Step 3: Simplify the Equation

Simplify the equation further to : \(4(x-3)^2+25(y+2)^2=116 \)

Step 4: Conversion to the Standard Form of Ellipse

The standard form of the ellipse is \((x-h)^{2}/a^{2} +(y-k)^{2}/b^{2} = 1\). Hence, by dividing by 116, the ellipse equation becomes: \((x-3)^2/29 + (y+2)^2/4.64 = 1\)

Step 5: Graphing the Ellipse

Now that we have the standard form of the ellipse, you can graph it with center at (3, -2), a= \(\sqrt{29}\) and b=\(\sqrt{4.64}\).

Step 6: Location of the Foci

Because a>b, we use the formula \(a=\sqrt{b^2 + c^2}\) to find c (the distance from center to each focus) which is found to be \(\sqrt{29-4.64}\)

Key Concepts

Completing the Square

Completing the square is an essential algebraic technique used to transform quadratic expressions into a perfect square trinomial. This transformation is crucial when working with conic sections like ellipses, as it allows us to rewrite the equation in a more useful form. Here's how completing the square works, focusing on a term such as \(4x^{2} - 24x\):

• First, factor out the coefficient of the square term, if necessary (in this case, \(4\) from \(4x^{2}\)), resulting in \(4(x^{2} - 6x)\).
• Next, find the number to complete the square. Take half of the coefficient of \(x\) (which is \(-6\)), giving \(-3\). Square it to get \(9\).
• Add \(9\) inside the parentheses and multiply it back out by the original factor, \(4\). In this case, the expression becomes \(4(x^{2} - 6x + 9)\).
• To keep the equation balanced, also add the product of the factor and the new number \(4 \times 9 = 36\) to both sides of the equation.

Repeat a similar process for the \(y\)-terms, ensuring the equation remains balanced. Completing the square for both \(x\) and \(y\) positions us to convert the quadratic equation of the ellipse into its standard form.

Standard Form of Ellipse

The standard form of an ellipse makes identifying its key properties much easier, such as its center, axes, and shape. Once an equation is converted into this form, the geometry of the ellipse becomes evident.The standard form 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.
• \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
• \(a\) is always the larger number when the ellipse is oriented horizontally, and can be easily identified as the variable with the larger denominator under the squared terms.
• \(b\) aligns with the smaller denominator.

For the given exercise, converting the equation into standard form involved dividing the entire equation by 116 to match the form \(1\) on the right side. The transformed equation is:\[\frac{(x-3)^2}{29} + \frac{(y+2)^2}{4.64} = 1\]This reveals the center at (3, -2), and gives clearer insight into the ellipse's dimensions with respect to \(x\) and \(y\).

Foci of Ellipse

The foci of an ellipse are two special points on the interior of the ellipse. The sum of the distances from any point on the ellipse to the foci is constant. Identifying these points helps in understanding the shape and properties of the ellipse.To determine the foci, knowing the lengths of the axes is crucial:

• If \(a > b\), the semi-major axis is along the x-direction, and the foci are located at \((h \pm c, k)\).
• If \(b > a\), the semi-major axis is along the y-direction, and the foci are located at \((h, k \pm c)\).

Here, use the formula,\[a = \sqrt{b^2 + c^2}\]Solving for \(c\) (the distance from the center to each focus), gives us \[c = \sqrt{29 - 4.64} \approx 4.59\]This calculation shows the foci are at \((3 \pm 4.59, -2)\). Knowing the location of the foci equips you with a deeper understanding of the ellipse's spatial extent and the nature of its curves.