Problem 20
Question
Identify the focus and the directrix of the graph of each equation. $$ y=\frac{1}{2} x^{2} $$
Step-by-Step Solution
Verified Answer
The focus of the parabola defined by the equation \(y=\frac{1}{2}x^{2}\) is at (0,2), and the directrix is the line \(y=-2\).
1Step 1: Identify the type and direction of the parabola
In the given equation \(y=\frac{1}{2}x^{2}\), the square term is on x. Therefore, this is a vertical parabola. The positive coefficient of \(x^{2}\) indicates that the parabola opens upwards.
2Step 2: Identify the vertex of the parabola
The equation does not contain any shifting (h,k), so the vertex of the parabola is the origin \((0,0)\).
3Step 3: Calculate the focus of the parabola
The value of 'a' in the equation is \(a = \frac{1}{2}\). The coordinates of the focus based on the vertex \((h,k)\) and the value of \(a\) in a vertical parabola that opens upwards are \((h,k+\frac{1}{4a})\). So the focus of the parabola is \((0,0+\frac{1}{4*\frac{1}{2}})=(0,2)\). So the focus of the given parabola is at (0,2).
4Step 4: Calculate the directrix of the parabola
For a parabola opening upwards or downwards, the equation of the directrix is \((y=k-\frac{1}{4a})\). Substituting the values, we get the equation of the directrix to be \((y=0-\frac{1}{4*\frac{1}{2}})\) or \(y=-2\).
Key Concepts
VertexFocusDirectrixVertical Parabola
Vertex
The vertex of a parabola is a crucial point because it serves as the "tip" or "turning point" of the parabola. In simple terms, it's the point where the parabola changes direction. For vertical parabolas, which open either up or down, the vertex determines the lowest or highest point respectively.
In the equation given, \( y = \frac{1}{2} x^2 \), there is no addition or subtraction term attached to \( x^2 \). This means there is no horizontal or vertical shift of the parabola.
In the equation given, \( y = \frac{1}{2} x^2 \), there is no addition or subtraction term attached to \( x^2 \). This means there is no horizontal or vertical shift of the parabola.
- Hence, the vertex is simply at the origin \((0,0)\).
Focus
The focus of a parabola is one of its defining features. It is a point from which distances are measured to construct the parabola. For vertical parabolas, the focus lies along the axis of symmetry.
Given the equation \( y = \frac{1}{2} x^2 \), we identify the value of \( a \) which is \( \frac{1}{2} \) in this instance. The formula to find the focus of a vertical parabola opening upwards is given by:
Given the equation \( y = \frac{1}{2} x^2 \), we identify the value of \( a \) which is \( \frac{1}{2} \) in this instance. The formula to find the focus of a vertical parabola opening upwards is given by:
- \((h, k + \frac{1}{4a})\)
- \(( 0, 0+ \frac{1}{4 \times \frac{1}{2}} ) = (0, 2)\)
Directrix
The directrix is a straight line that is part of the geometric definition of a parabola. For vertical parabolas, this line is horizontal and sits opposite to the direction in which the parabola opens.
For the given parabola \( y = \frac{1}{2} x^2 \), the directrix can be found using the formula:
For the given parabola \( y = \frac{1}{2} x^2 \), the directrix can be found using the formula:
- \( y = k - \frac{1}{4a} \)
- \( y = 0 - \frac{1}{4 \times \frac{1}{2}} = -2 \)
Vertical Parabola
A vertical parabola is one that opens either upwards or downwards, depending on the sign of the \( a \) term in the equation \( y = ax^2 \). With a positive \( a \), like in the equation \( y = \frac{1}{2} x^2 \), the parabola opens upwards.
Key characteristics of a vertical parabola include:
Key characteristics of a vertical parabola include:
- The vertex as the lowest or highest point.
- The axis of symmetry, a vertical line through the vertex.
- The focus sitting along this axis of symmetry.
- The directrix being a horizontal line perpendicular to the axis of symmetry.
Other exercises in this chapter
Problem 20
Find the foci for each equation of an ellipse. Then graph the ellipse. $$ \frac{x^{2}}{81}+\frac{y^{2}}{49}=1 $$
View solution Problem 20
Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center
View solution Problem 20
For each equation, find the center and radius of the circle. $$ (x+2)^{2}+(y-10)^{2}=4 $$
View solution Problem 21
Find the foci for each equation of an ellipse. Then graph the ellipse. $$ \frac{x^{2}}{100}+\frac{y^{2}}{36}=1 $$
View solution