Problem 23 - College Algebra

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. $$ y^{2}+12 x-6 y+21=0 $$

Step-by-Step Solution

Answer

Vertex: (1, 3); Focus: (-2, 3); Directrix: x = 4.

Step 1: Rearrange and Group Terms

Start with the given equation: \[ y^2 + 12x - 6y + 21 = 0 \]Group the terms involving \( y \) on one side:\[ y^2 - 6y = -12x - 21 \]

Step 2: Complete the Square

To complete the square for the terms in \( y \), take half of the linear coefficient of \( y \), square it, and add it to both sides of the equation. Here, the linear coefficient is \(-6\), and half of it is \(-3\). Squaring gives \(9\). Add \(9\) to both sides:\[ y^2 - 6y + 9 = -12x - 21 + 9 \]This simplifies to:\[ (y - 3)^2 = -12x - 12 \]

Step 3: Rearrange to Standard Form

Factor the right-hand side of the equation:\[ (y - 3)^2 = -12(x - 1) \]This is the standard form of a vertical parabola, \[ (y - k)^2 = 4p(x - h) \] where the vertex is \((h, k)\) and \(4p\) is the coefficient of \((x-h)\).

Step 4: Determine Vertex, Focus, and Directrix

From the standard form \((y - 3)^2 = -12(x - 1)\), identify that \((h, k) = (1, 3)\). So, the vertex \(V\) is \((1, 3)\).Since \(4p = -12\), we have \(p = -3\). For a vertical parabola that opens to the left, the focus \((F)\) is \((h + p, k) = (1 - 3, 3) = (-2, 3)\).The directrix \((d)\) is the vertical line given by \(x = h - p = 1 + 3 = 4\).

Key Concepts

Standard Form

The standard form of a parabola is a way of expressing its equation so that it clearly defines the shape and position of the parabola. There are different forms depending on the orientation of the parabola (opening upwards, downwards, left, or right). For the given equation,

• The standard form is \[ (y - k)^2 = 4p(x - h) \]for a parabola opening left or right.
• If the parabola opens upwards or downwards, the form is \[ (x - h)^2 = 4p(y - k) \].

This standardization simplifies the process of finding the vertex, focus, and directrix, making it easier to graph the parabola.
In our exercise, after rearranging and completing the square, the equation is \[ (y - 3)^2 = -12(x - 1) \].
This is the standard form for a parabola opening to the left.

Vertex

The vertex of a parabola is a critical point that defines the turning point or the peak of the parabola. It's the point where the parabola changes its direction.

• Using the standard form \[ (y - k)^2 = 4p(x - h) \]
• The vertex \((h, k)\) is directly given by the values of \(h\) and \(k\) within the equation.

For the parabola in the exercise, the vertex is at \((1, 3)\).
This point is the midpoint between the focus and the directrix, and it is crucial in determining the orientation and position of the parabola in the Cartesian plane.

Focus

The focus of a parabola is a point that, along with the directrix, helps to define the shape and orientation of the parabola. It is one of the focal points from which the distance to any point on the parabola is the same as the distance from that point to the directrix.

• In the standard form \[ (y - k)^2 = 4p(x - h) \]
• Focus is defined as the point \((h + p, k)\).

For our given parabola, since \(4p = -12\) which implies \(p = -3\), the focus is located at \((-2, 3)\).
The focus lies on the axis of symmetry and helps in tracing out the parabola graph.

Directrix

A parabola's directrix is a line that, together with the focus, guides its shape by maintaining the distance relationship of any point on the parabola to the focus and the directrix.

• In the case of a vertical parabola \[ (y - k)^2 = 4p(x - h) \]
• The directrix is a vertical line given by \(x = h - p\).

In this exercise, it is located at \(x = 4\).
This line is perpendicular to the axis of symmetry and doesn't intersect with the parabola. Its relationship with the focus plays a vital role in forming the curve of the parabola consistently.