Problem 74 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the vertex, focus, and directrix of the parabola and sketch its graph. Use a graphing utility to verify your graph. $$x^{2}-2 x+8 y+9=0$$

Step-by-Step Solution

Answer

The vertex of the parabola is (1,1), the focus is at (1,3) and the equation of the directrix is \(y=-1\). The parabola opens downwards.

Step 1: Conversion to standard form

Looking at the equation \(x^{2}-2 x+8 y+9=0\), rearrange it into the form \((x-h)^2=4a(y-k)\) by completing the square if necessary. After simplification, we find \((x-1)^2 = -8(y-1)\), which is the standard form of the equation.

Step 2: Identify the vertex

The vertex of the parabola is the point \((h,k)\). From the standard form equation, compare it to \((x-h)^2 = 4a(y-k)\) to find \((h,k)=(1,1)\). So, the vertex of the parabola is at the point (1, 1).

Step 3: Determine the value of 'a'

From the standard form \((x-1)^2 = -8(y-1)\), we see that \(4a=-8\). Solving for \(a\) gives \(a=-2\). This value is useful in finding the focus and the directrix of the parabola.

Step 4: Calculate the focus

The coordinates of the focus are \((h,k-a)\), so in this case, the focus equals \((1, 1+2) = (1,3)\).

Step 5: Find the equation of the directrix

The equation of the directrix is \(y=k+a\). With \(k=1\) and a = -2, the directrix becomes \(y=1-2 = -1\).

Step 6: Graph the parabola

Begin your sketch by marking the vertex (1,1) and drawing the parabola opening downwards due to \(a<0\). Mark the focus point (1,3), and note that it is 2 units above the vertex, and draw the directrix as a horizontal line at \(y=-1\).

Key Concepts

Vertex

The vertex of a parabola is a crucial point that serves as the "turning point" of the curve. For the equation given in the exercise, we start by rewriting it in the standard form of a parabola,

• Standard form: \( (x-h)^2 = 4a(y-k) \)

After simplifying the original equation \(x^2 - 2x + 8y + 9 = 0\) using completing the square, we transform it into \( (x-1)^2 = -8(y-1) \).
Here, the vertex is represented by the point \((h, k)\), which means our

• Vertex: \((1, 1)\)

The vertex tells us where the parabola changes direction and determines the symmetry line of the parabola. In this case, since the parabola opens downwards, this vertex is also the maximum point on the graph.
The role of the vertex is pivotal because it simultaneously serves as a reference point for finding the focus and the directrix.

Focus

The focus of a parabola is a point that plays a significant role in the geometric definition and creation of the parabola. The focus lies inside the curve, and every point on the parabola is equidistant from the focus and the directrix.

• Focus Formula: \((h, k - a)\)

In the equation \( (x-1)^2 = -8(y-1) \), we found that \(a = -2\).
So calculating the focus gives us:

• Focus: \((1, 3)\)

This focus coordinates \((1, 3)\) translate to being 2 units above the vertex in this context. The orientation of the parabola is determined by the negative \(a\) value, pointing downwards.
The focus is fundamental for drawing because it helps in manually sketching the parabola using the symmetry and various point tests from the focus-drawn rays.

Directrix

The directrix of a parabola is an equally important concept in the formation of the parabola. The directrix is a line that is perpendicular to the axis of symmetry of the parabola. The set of points on the parabola have equal distances to the directrix and the focus.

• Directrix Formula: \(y = k + a\)

In our equation's case, with the vertex \( (1, 1) \) and \(a = -2\), we compute the directrix as:

• Directrix: \(y = -1\)

The directrix serves as a guide line at \(y = -1\), which is parallel to the x-axis.
This line is integral in ensuring the distance ratio required by each parabola is satisfied, as every point on the parabola maintains equal distance to both the focus at \((1, 3)\) and the line \(y = -1\).
Understanding the directrix gives insight into the behavior and symmetry nature of the parabola and ensures great accuracy when sketching it.