Problem 43 - Precalculus

Question

Find the vertex, focus, and directrix of the parabola. Use a graphing utility to graph the parabola. $$x^{2}+4 x+6 y-2=0$$

Step-by-Step Solution

Answer

The vertex of the parabola is (-2, 1/3), the focus is (-2, 5/6), and the equation of the directrix is y = -4/3.

Step 1: Rewrite the equation in standard form

The standard form of a parabola equation which opens upwards or downwards is \( (x-h)^2 = 4p(y-k) \), where (h,k) is the vertex. Rewrite the given equation \(x^{2}+4 x+6 y-2=0\) to the form \(x^2 = 4p(y-k)\). We complete the square for the x-terms to achieve this. So we have \( (x+2)^2 -4 + 6y -2 = 0\), which simplifies to \( (x+2)^2 = 6y - 2\), hence \( (x+2)^2 = 6(y - 1/3)\).

Step 2: Identify the vertex

From the standard form obtained in step 1: \( (x+2)^2 = 6(y - 1/3)\), we see the vertex (h,k) is (-2, 1/3).

Step 3: Find the value of p and identify the focus

From the standard form we know that 4p = 6, therefore p = 3/2. Since this parabola opens upwards, the focus will be (h, k+p) = (-2, 1/3 + 3/2) = (-2, 5/6)

Step 4: Identify the directrix

The directrix of a parabola that opens upwards or downwards is given by y = k-p. Hence the directrix is y = 1/3 - 3/2 = -4/3.

Key Concepts

Vertex

The vertex of a parabola is a crucial point located right at the curve where the parabola changes direction.
It acts as a sort of 'turning point' or 'tip' of the parabola.
In mathematical terms, the vertex is represented as a point \(h, k\).

For our equation \(x^{2}+4x+6y-2=0\), we first rewrite it in a form that reveals the vertex.
By completing the square, we expressed it as \((x+2)^2=6(y-1/3)\).
This new form tells us that the vertex is at \((-2, 1/3)\).
What this means is:

• The parabola has its turning point at x = -2 and y = 1/3.
• From the graph, you can see this as the point where the sharpest direction change occurs.

In practical terms, identifying the vertex helps in graphing, as you know exactly where to "pin" one part of the parabola.
This makes the remaining points follow around this central position.

Focus

The focus of a parabola plays a vital role in its geometric shape.
It lies on the axis of symmetry of the parabola and is a fixed point that helps define its properties.
In simple terms, the focus can be thought of as the point towards which the parabola curves.

For our equation restructured to \((x+2)^2=6(y-1/3)\), the focus tells us something critical about the curve's behavior.
You find it by calculating \(h, k + p\), where \(p\) is the distance between the vertex and the focus.
Here, since \(4p\) equals 6, \((p = \frac{3}{2})\).
Thus, the focus is at \((-2, \frac{5}{6})\).
This position tells us a few things:

• It's above the vertex since \(p\) is added to the vertex's y-coordinate for upward-opening parabolas.
• All lines (or radii) drawn from any point on the parabola to the focus will be equal in length when combined with the perpendicular distance to the directrix.

This information is instrumental in determining how the parabola opens by showing its curvature's point of attraction.

Directrix

A directrix is a straight line utilized alongside the focus to provide a precise definition of a parabola.
It works in tandem with the focus to guide the parabolic curve.
Think of the directrix as a baseline from which distances are evaluated.

For our parabola in the form \((x+2)^2=6(y-1/3)\), you find its directrix using the formula \(y=k-p\).
Here, we know \(p=\frac{3}{2}\), so the directrix is \((y = \frac{1}{3} - \frac{3}{2})\), which simplifies to \(y = -\frac{4}{3}\).
This insight into the directrix provides us with:

• Confirmation that the parabola opens upwards since the directrix is a straight line below the vertex.
• Assurance of equal distances between any point on the parabola to the directrix and to the focus.

The directrix acts like a guiding line that helps to maintain the shape of the parabola, ensuring it's symmetric around its axis.