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Magazine Chapter 10: Problem 6
Find the vertex, focus, and directrix of the parabola, and sketch the graph. $$(y+5)^{2}=-6 x+12$$
Short Answer
Expert verified
Vertex: (2, -5); Focus: (1/2, -5); Directrix: x = 7/2.
Step by step solution
01
Rewrite the Equation in Standard Form
The given equation is \((y+5)^2 = -6x + 12\). Start by isolating the variable terms involving \(x\) on one side. Therefore, rewrite it as:\[(y+5)^2 = -6(x-2)\] This formula shows a parabola of the form \((y-k)^2 = 4p(x-h)\) where \((h, k)\) is the vertex and \(p\) is the parameter that helps find the focus and directrix.
02
Identify the Vertex
From the standard form of the equation \((y-k)^2 = 4p(x-h)\), we identify the vertex as \((h, k)\). Comparing terms with the rewritten equation \((y+5)^2 = -6(x-2)\), we find:- \(h = 2\)- \(k = -5\)Thus, the vertex is \((2, -5)\).
03
Determine the Parameter p
The equation \((y+5)^2 = -6(x-2)\) can be compared to the standard form \(4p(x-h)\). Here, \(-6\) corresponds to \(4p\). Solving \(4p = -6\), we find:- \(p = -\frac{3}{2}\).This indicates that the parabola opens to the left.
04
Find the Focus
The focus of a parabola is at \((h + p, k)\). Given that \(h = 2\), \(k = -5\), and \(p = -\frac{3}{2}\), we substitute these into the formula:- Focus is \((2 - \frac{3}{2}, -5) = (\frac{1}{2}, -5)\).
05
Determine the Directrix
The directrix of a parabola \((y-k)^2 = 4p(x-h)\) is given by \(x = h - p\). Substituting the values of \(h\) and \(p\):- Directrix: \(x = 2 + \frac{3}{2} = \frac{7}{2}\).
06
Sketch the Graph
To sketch the parabola, plot the vertex \((2, -5)\), the focus \((\frac{1}{2}, -5)\), and draw the line \(x = \frac{7}{2}\) for the directrix. Since \(p\) is negative, the parabola opens to the left, centered at the vertex, curving around the focus and moving away from the directrix.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex
The vertex of a parabola is a crucial point as it indicates the turning point of the curve. It's where the parabola changes direction. In a parabola equation of the form \((y-k)^2 = 4p(x-h)\), the vertex is represented by the coordinates \((h, k)\).
- In this form, \(h\) moves the parabola left or right, and \(k\) moves it up or down.
- As the vertex is the point of symmetry, it helps in sketching the parabola accurately.
Focus
The focus of a parabola is a point inside the curve that helps in defining its shape. Every point on a parabola is equidistant from the focus and the directrix. Hence, the focus significantly affects the parabola's curvature.To find it using the equation \((y-k)^2 = 4p(x-h)\), we apply \(h + p\) for the x-coordinate and \(k\) for the y-coordinate.
- If \(p\) is positive, the parabola opens to the right, and if \(p\) is negative, it opens to the left.
Directrix
The directrix is a line that, together with the focus, helps to define the parabola. It lies outside the curve and is perpendicular to the axis of symmetry, serving as a reference point.The general formula for finding the directrix of a sideways parabola \((y-k)^2 = 4p(x-h)\) is \(x = h - p\). In this particular problem:
- The directrix provides a line perpendicular to the direction in which the parabola opens.
- Just like the focus, it is equidistant to all points on the parabola.
