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Magazine Chapter 11: Problem 8
Exer. 1-12: Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$ (y+1)^{2}=-12(x+2) $$
Short Answer
Expert verified
Vertex: (-2, -1); Focus: (-5, -1); Directrix: x = 1.
Step by step solution
01
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 \(4p = -12\).
02
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.
03
Determine the Value of \(p\)
Using the equation \(4p = -12\), solve for \(p\): \(p = \frac{-12}{4} = -3\).
04
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: \((-2 + (-3), -1) = (-5, -1)\).
05
Determine the Directrix
The directrix of a horizontal parabola is the vertical line \(x = h - p\). Here, \(x = -2 - (-3) = 1\).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(k = -1\).
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 \((-5, -1)\).
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 \(x = 1\).
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鈥檚 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鈥檛 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.
