Problem 56
Question
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(0,0), \text { Endpoints }(2,1),(-2,1) $$
Step-by-Step Solution
Verified Answer
The equation of the parabola is \( y = \frac{1}{4}x^2 \).
1Step 1: Understand the Properties of the Parabola
The vertex of a parabola is given by \( V(0,0) \), and the endpoints of the latus rectum are \( (2,1) \) and \( (-2,1) \). Since both endpoints have the same \( y \)-coordinate, the parabola must be vertically oriented.
2Step 2: Determine the Axis of Symmetry and Direction
Since the vertex \( V(0,0) \) is at the origin and the latus rectum is horizontal, the parabola opens upwards or downwards. Given that the \( y \)-coordinates of both endpoints are above the vertex, the parabola opens upwards.
3Step 3: Calculate the Length of the Latus Rectum
Calculate the distance between the endpoints of the latus rectum. The endpoints are \( (2,1) \) and \( (-2,1) \). This distance is \( |2 - (-2)| = 4 \). Thus, the length of the latus rectum is 4.
4Step 4: Determine the Value of \( a \)
For the equation \( y = ax^2 \) of a vertically oriented parabola, the formula for the length of the latus rectum is \( \frac{1}{|a|} = 4 \). Therefore, \(|a| = \frac{1}{4}\). Since it opens upwards, \( a = \frac{1}{4} \).
5Step 5: Write the Equation of the Parabola
The equation of the parabola with vertex at the origin is given by \( y = ax^2 \). Using \( a = \frac{1}{4} \), the equation becomes \( y = \frac{1}{4}x^2 \).
Key Concepts
VertexLatus RectumAxis of SymmetryEquation of Parabola
Vertex
The vertex of a parabola is like its nose, the pointy tip where it changes direction. Think of it as the most prominent point. In this exercise, the vertex is at the origin of our coordinate system, which is the point \( (0,0) \). The vertex gives us a clear starting point. It is essential for forming the equation of the parabola. If you imagine drawing a smile with a pencil, the vertex is where you first touch the paper.
- The vertex helps to define the orientation of the parabola.
- It is used to find the equation of the parabola.
- In this case, because our vertex is at \( (0,0) \), it simplifies many calculations.
Latus Rectum
The latus rectum connects points directly related to the parabola's width. In simple terms, it's a line segment that passes through the focus of the parabola and is perpendicular to the axis of symmetry. In this exercise, the endpoints of the latus rectum are \( (2,1) \) and \( (-2,1) \). These coordinates help in determining how wide or narrow the parabola is.
- The length of the latus rectum is directly related to the value of \( a \) in the parabola's equation.
- The calculated distance here was 4, which tells us about the parabola's width.
Axis of Symmetry
The axis of symmetry is like the imaginary line that divides the parabola into two mirror-image halves. For this exercise, the parabola is vertically oriented, making the axis of symmetry a horizontal line. Since our vertex is at the origin \( (0,0) \), and the parabola opens upwards, the axis is simply the \( y-\)axis.
- Every parabola has one axis of symmetry.
- This axis helps us understand the parabola's orientation.
Equation of Parabola
The equation of a parabola is like its mathematical signature. For a vertically oriented parabola with vertex at the origin, the equation can be expressed as \( y = ax^2 \). In this particular exercise, we deduced that \( a = \frac{1}{4} \), giving us the complete equation \( y = \frac{1}{4}x^2 \).This equation tells us several things:
- The overall shape of the parabola.
- Its orientation - whether it opens upwards or downwards, which in this case is upwards.
Other exercises in this chapter
Problem 56
Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form
View solution Problem 56
Express each equation in polar form with \(r\) as a function of \(\theta\). $$ x y=2 $$
View solution Problem 56
For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x .\) Express as simply as possible. Use a g
View solution Problem 56
For the following exercises, determine the value of \(k\) based on the given equation. Given \(4 x^{2}+k x y+16 y^{2}+8 x+24 y-48=0\) find \(k\) for the graph t
View solution