Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-12: Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$ (y+1)^{2}=-12(x+2) $$

Step-by-Step Solution

Answer

Vertex: (-2, -1); Focus: (-5, -1); Directrix: x = 1.

Step 1: Identify the Form of the Parabola

The given equation \((y+1)^2 = -12(x+2)\) is in the form \((y-k)^2 = 4p(x-h)\), which represents a horizontal parabola. Here, \(h = -2\), \(k = -1\), and \(4p = -12\).

Step 2: Find the Vertex

The vertex \((h, k)\) of the parabola is \((-2, -1)\). This is obtained directly from the given equation by recognizing the vertex form structure.

Step 3: Determine the Value of \(p\)

Using the equation \(4p = -12\), solve for \(p\): \(p = \frac{-12}{4} = -3\).

Step 4: Find the Focus

For a horizontal parabola opening left, the focus is \((h + p, k)\). Substitute \(h = -2\), \(k = -1\), and \(p = -3\) to get the focus: \((-2 + (-3), -1) = (-5, -1)\).

Step 5: Determine the Directrix

The directrix of a horizontal parabola is the vertical line \(x = h - p\). Here, \(x = -2 - (-3) = 1\).

Step 6: Sketch the Graph

Draw the parabola with its vertex at \((-2, -1)\). Plot the focus at \((-5, -1)\) and draw the directrix line \(x = 1\). Ensure the parabola opens to the left, consistent with the negative \(p\) value.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key point that helps us understand the shape and position of the parabola on the graph. In the context of the given equation \((y+1)^2 = -12(x+2)\), the vertex is found by identifying the values of \(h\) and \(k\).

• The vertex represents the point \((h, k)\), which makes it an extremely significant point because it's the 'tip' or the 'turning point' from where the parabola changes direction.
• For our equation, the vertex is located at \((-2, -1)\) which you can directly gather from the structure of the vertex form \((y-k)^2 = 4p(x-h)\). In this equation, \(h = -2\) and \(k = -1\).

The vertex can help determine how to draw the basic shape of the parabola on the coordinate plane. It's crucial for understanding whether the parabola opens right, left, upward, or downward. In our context, since the parabola opens horizontally, this vertex is where it begins to open to the left as indicated by the subsequent calculations.

Focus of a Parabola

The focus of a parabola is another important component when studying its geometric properties. The focus lies on the axis of symmetry of the parabola and helps in determining its shape.

• For horizontal parabolas, like the one given in \((y+1)^2 = -12(x+2)\),the focus is calculated using the vertex form parameters and the derived value of \(p\).
• First, we solve from the equation \(4p = -12\) to get \(p = -3\). With this, we can find the focus at \((h + p, k)\), thereby getting \((-5, -1)\).

This focus point helps us make sense of how the parabola is 'pulled' towards it, always staying equidistant from it and the directrix.In practical terms, the coordinate of the focus captures how concave or convex the parabola appears as it opens either to the left or right.

Directrix of a Parabola

The directrix of a parabola is a line that aids in understanding the width and opening of the parabola with respect to the focus. It serves as a base reference line that helps define the parabola's path.

• The directrix works in tandem with the focus, such that a parabola is defined as the set of points equidistant from both the focus and the directrix.
• In our horizontal parabola, with the vertex at \((-2, -1)\)and p = -3, we find the directrix using the formula \(x = h - p\), resulting in \(x = 1\).

This line, placed vertically, complements the leftward opening of the parabola as the curve maintains equal distance from this line and the focus during its formation.Marking the directrix on your sketch guarantees accuracy when graphing parabolas, revealing how the structure is influenced by distance rules.

Graphing Parabolas

Graphing parabolas, especially when you know the vertex, focus, and directrix, makes for an easier, structured way of drawing these curved shapes. A parabola’s shape and orientation can change dramatically based on these guide points.

• Start by plotting the vertex of the parabola in our graph, which in this case is \((-2, -1)\), as it provides the turning point.
• Next, accurately mark the focus at \((-5, -1)\), which was calculated based on the negative \(p\), indicating the leftward openness.
• Don’t forget the directrix, which in this parabola's situation is the vertical line \(x = 1\). This line helps guide the stretch and direction of the parabola's curve.

By sketching these elements in coordination, you build a clearer, more detailed graph that reflects the parabola's orientation and symmetry.Graphing becomes much simpler when you have these markers acting as a framework for where the curve should go.Tips like identifying the correct opening (left) due to the negative coefficient, and using the vertex-focus to direct the parabolic path, ensure a precise representation.