Problem 11 - Intermediate Algebra

Question

Determine whether the graph of each equation is a circle or a parabola. A. \(x^{2}+y^{2}-6 x+8 y-10=0\) B. \(y^{2}-2 x+3 y-9=0\) C. \(x^{2}+5 x-y=0\) D. \(x^{2}+12 x+y^{2}=0\)

Step-by-Step Solution

Answer

A is a circle, B and C are parabolas, and D is invalid as a circle.

Step 1: Identify Terms for Equation A

The equation given is \(x^{2}+y^{2}-6x+8y-10=0\). By observing this, we note that both \(x^2\) and \(y^2\) are present, and their coefficients are equal, which is typical for a circle. To confirm, we can rewrite it in standard form by completing the square.

Step 2: Complete the Square for Equation A

To complete the square, we rearrange it as \((x^2 - 6x) + (y^2 + 8y) = 10\).For \(x\), complete the square: Take half of \(-6\), square it, and add it inside: \((x-3)^2-9\).For \(y\), take half of \(8\), square it, and add it inside: \((y+4)^2-16\).Now, add the inverse constants to 10: \((x-3)^2 + (y+4)^2 = 25\).This confirms it is a circle with center \((3, -4)\) and radius 5.

Step 3: Identify Terms for Equation B

The equation given is \(y^2-2x+3y-9=0\). This equation contains the term \(y^2\) only, with no \(x^2\) term, which is characteristic of a parabola opening along the x-axis.

Step 4: Rearrange Equation B into Standard Form

Rearrange terms so all \(y\) terms are on one side and the \(x\) term on the other: \(y^2 + 3y = 2x + 9\).Thus, the equation takes the form of a parabola.

Step 5: Identify Terms for Equation C

The equation given is \(x^2 + 5x - y = 0\). This equation contains the term \(x^2\) and \(y\), but no \(y^2\), which indicates that it is a parabola.

Step 6: Rearrange Equation C into Standard Form

Move \(y\) to the other side to emphasize the parabola form: \(x^2 + 5x = y\).Thus, this represents a parabola.

Step 7: Identify Terms for Equation D

The equation given is \(x^2 + 12x + y^2 = 0\). Both \(x^2\) and \(y^2\) terms are present and have the same coefficient, suggesting it would be a circle. However, the constant term being zero suggests an error, as no geometric figure exists on a purely imaginary plane in standard contexts in basic geometry. Thus, this is an invalid equation for a circle.

Key Concepts

Circle

When dealing with equations, identifying a circle is usually straightforward once you recognize the presence of both squared terms, like \(x^2\) and \(y^2\). In the standard form of a circle’s equation, both these terms have the same coefficient, typically expressed as:

\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here,

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

Completing the square is a key algebraic technique used to convert an equation into this recognizable form, which helps distinguish circles from other graphs.
In exercise A, after completing the square for both \(x\) and \(y\), the equation becomes \((x-3)^2 + (y+4)^2 = 25\). This represents a circle centered at \((3, -4)\) with a radius of 5.

Parabola

A parabola typically involves either an \(x^2\) term or a \(y^2\) term, but not both as squared terms. The general form for a parabola aligned with the axes is:

- Opens along the x-axis: \((y-k)^2 = 4p(x-h)\) - Opens along the y-axis: \((x-h)^2 = 4p(y-k)\)
In these expressions,

• \(h, k\) represent the vertex of the parabola.
• \(p\) is the focal length from the vertex to the focus of the parabola.

For instance, in exercise B: \(y^2 - 2x + 3y - 9 = 0\), the lack of an \(x^2\) term coupled with the square in \(y\) indicates it is a parabola. Rearranging confirms this by isolating \(y\) terms, leading to \(y^2 + 3y = 2x + 9\). Similarly, in exercise C, \(x^2 + 5x = y\) directly shows a parabola opening upwards along the y-axis.

Completing the Square

Completing the square is an essential algebra technique used to rewrite quadratic equations, making it easier to recognize forms such as circles or parabolas. The process involves:

• Identifying and isolating the quadratic term(s), for example, \(x^2\) or \(y^2\).
• Taking half the coefficient of the linear term (like \(bx\) or \(cy\)), squaring it, and then adding and subtracting that square.
• Rewriting as a perfect square trinomial.

This technique is evident in exercise A with \(x^2 + y^2 - 6x + 8y - 10 = 0\). By completing the square for \(x\) with \(x^2 - 6x\) and \(y\) with \(y^2 + 8y\), we transform the equation into a standard form for a circle: \((x-3)^2 + (y+4)^2 = 25\). This provides clarity and confirms the type of graph represented by the equation. Understanding this method is invaluable for recognizing and interpreting various quadratic graphs.