Chapter 10: Problem 7
Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. $$6 y-2 x^{2}=0$$
Short Answer
Expert verified
Focus: (0, 3/4); Directrix: y = -3/4.
Step by step solution
01
Rewrite the Equation
First, rewrite the given equation in the standard form of a parabola that opens vertically. The given equation is: \( 6y - 2x^2 = 0 \).Rearrange the equation to isolate \( y \) on one side:\( 6y = 2x^2 \)Divide everything by 6 to simplify:\( y = \frac{1}{3}x^2 \)
02
Identify Parabola Parameters
The standard form of a parabola that opens vertically is \( y = ax^2 \). Comparing to our equation \( y = \frac{1}{3}x^2 \), we identify that \( a = \frac{1}{3} \).
03
Determine the Focus
The focus of a parabola \( y = ax^2 \) is located at \( \left(0, \frac{1}{4a}\right) \).Substitute \( a = \frac{1}{3} \) into the formula:\( \frac{1}{4a} = \frac{1}{(4 \times \frac{1}{3})} = \frac{3}{4} \).Thus, the coordinates of the focus are \( (0, \frac{3}{4}) \).
04
Find the Equation of the Directrix
The equation of the directrix for a parabola \( y = ax^2 \) is given by \( y = -\frac{1}{4a} \).Using \( a = \frac{1}{3} \), calculate:\( y = -\frac{1}{4 \times \frac{1}{3}} = -\frac{3}{4} \).Thus, the equation of the directrix is \( y = -\frac{3}{4} \).
05
Sketch the Parabola
On a coordinate plane, plot the vertex of the parabola at the origin \((0,0)\), since the parabola is \( y = \frac{1}{3}x^2 \). Add the focus \((0, \frac{3}{4})\) above the vertex and draw the directrix line at \( y = -\frac{3}{4} \).Sketch the U-shaped curve that opens upwards, ensuring it extends equally on both sides of the y-axis, passing through the vertex.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Focus of a Parabola
The focus of a parabola is a key point that defines the shape and direction of the parabola itself. In the standard form of a vertically opening parabola, which is given by the equation \( y = ax^2 \), the focus can be found using the formula \( \left(0, \frac{1}{4a}\right) \). This focus point is located along the axis of symmetry, perpendicular to the directrix.
- The focus is always inside the parabola.
- It helps guide the "U" shape of the curve.
- The distance from the vertex to the focus is equal to the distance from the vertex to the directrix.
Directrix
The directrix is a straight line associated with a parabola and is vital for defining its geometry. It is located opposite the focus in relation to the vertex. For a vertical parabola with the equation \( y = ax^2 \), the directrix can be calculated using the formula \( y = -\frac{1}{4a} \). This line provides a boundary that the parabola "bends away" from.
- The directrix helps in determining the width and orientation of the parabola.
- It is exactly the same distance from the vertex as the focus is, but in the opposite direction.
- It does not intersect with the parabola itself.
Vertex Form
The vertex form of a parabola is a way of expressing the quadratic equation that makes its features easier to discern. It is given by the formula \( y = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex coordinates. However, when the vertex is at the origin, \( (0, 0) \), the equation simplifies to \( y = ax^2 \), like in our problem.
- The vertex form is extremely useful for instantly identifying the vertex of the parabola.
- It helps in quickly determining direction (upward or downward) based on the sign of \( a \).
- It allows for easy calculation of important points such as the focus and the position of the directrix.