Problem 29

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. \(3 y^{2}-4 x-6 y+23=0\)

Step-by-Step Solution

Verified

Answer

The vertex is (5, 1), focus is (16/3, 1), and directrix is x = 14/3.

1Step 1: Rewrite the equation in standard form

Start with the given equation: \[3y^2 - 4x - 6y + 23 = 0\]Rearrange terms to isolate the terms involving "y" on one side:\[3y^2 - 6y = 4x - 23\]Complete the square for the terms involving "y". First, factor out the 3:\[3(y^2 - 2y) = 4x - 23\]To complete the square inside the parenthesis, add and subtract \(\left(\frac{-2}{2}\right)^2 = 1\):\[3(y^2 - 2y + 1 - 1) = 4x - 23\]This becomes:\[3((y - 1)^2 - 1) = 4x - 23\]Simplify to:\[3(y - 1)^2 - 3 = 4x - 23\]Add 3 to both sides:\[3(y - 1)^2 = 4x - 20\]Now, divide the entire equation by 4:\[ \frac{3}{4}(y - 1)^2 = x - 5\]Thus, the equation in standard form is:\[ (y - 1)^2 = \frac{4}{3}(x - 5) \]

2Step 2: Identify the vertex, focus, and directrix

The standard form of a parabola that opens sideways is \((y - k)^2 = 4p(x - h)\), where \((h, k)\) is the vertex.Comparing with our equation, \((y - 1)^2 = \frac{4}{3}(x - 5)\), we see:- The vertex \(V\) is \((5, 1)\).- \(4p = \frac{4}{3}\) implies \(p = \frac{1}{3}\).Since the parabola opens to the right (positive \(p\)), the focus \(F\) is \(\left(5 + \frac{1}{3}, 1\right) = \left(\frac{16}{3}, 1\right)\).The directrix is a vertical line \(x = h - p\). Thus:\(x = 5 - \frac{1}{3} = \frac{14}{3}\).

Key Concepts

Standard FormVertexFocusDirectrix

Standard Form

The standard form of a parabola's equation helps you easily identify key characteristics such as its geometric inclination and specific attributes. For a parabola that opens sideways, its standard form is given by:

\[(y - k)^2 = 4p(x - h)\]

Here, \((h, k)\) represents the vertex of the parabola, and \(p\) defines the distance from the vertex to both the focus and the directrix.
In the given problem, the initial equation was rearranged and rewritten to match this format, resulting in:

\[(y - 1)^2 = \frac{4}{3}(x - 5)\]

By placing the equation in standard form, not only can we determine the vertex but also the focus and directrix directly, providing a helpful overview of the parabola's structure.

Vertex

The vertex of a parabola is a crucial point. It signifies the axis of symmetry and tells us where the parabola changes its direction. In the standard form of a parabola

\[(y - k)^2 = 4p(x - h)\]

The expression \((h, k)\) gives the vertex directly.
For the exercise, comparison with the standard form

\[(y - 1)^2 = \frac{4}{3}(x - 5)\]

Revealed that the vertex is at

\(V = (5, 1)\)

Identifying the vertex is important as it serves as a reference point when graphing and analyzing the properties of the parabola.

Focus

The focus of a parabola is a special point that every point on the parabola reflects to equally. It is important because it indicates the direction and width of the parabola.
To find the focus, we need the value of \(p\) from the standard form:

\[(y - 1)^2 = \frac{4}{3}(x - 5)\]

Here, it was established that \(4p = \frac{4}{3}\), leading to \(p = \frac{1}{3}\). Therefore, the focus, partially defined as

\(F = (h+p, k)\)

Is at

\(F = \left( 5 + \frac{1}{3}, 1 \right) = \left( \frac{16}{3}, 1 \right) \)

Understanding the location of the focus is pivotal for drawing the parabola correctly and comprehending its physical properties.

Directrix

The directrix of a parabola is a line with which each point on the parabola maintains equal distance to the focus. This geometric property enhances the understanding of a parabola's shape and trajectory.
When using the standard form: