Problem 18
Question
Find the equation of the parabola whose vertex is the origin and whose axis is the \(y\)-axis if the parabola passes through the point \((-3,5)\). Make a sketch.
Step-by-Step Solution
Verified Answer
The parabola is \( y = \frac{5}{9}x^2 \).
1Step 1: Understanding the Vertex Form Equation
Since the vertex is at the origin and the axis is the y-axis, the equation of the parabola in vertex form is \( y = ax^2 \). This equation represents a parabola that opens upwards or downwards depending on the value of \(a\).
2Step 2: Substitute the Given Point into the Equation
The parabola passes through the point \((-3, 5)\). Substitute \(x = -3\) and \(y = 5\) into the equation \(y = ax^2\): \[ 5 = a(-3)^2 \].
3Step 3: Solving for the Coefficient a
Simplify the equation to find \(a\): \[ 5 = 9a \].Divide both sides by 9: \( a = \frac{5}{9} \).
4Step 4: Write the Equation of the Parabola
Now that we have found the value of \(a\), we can write the equation of the parabola as:\[ y = \frac{5}{9}x^2 \].
5Step 5: Sketching the Parabola
To sketch the parabola, note that it opens upwards since \(a > 0\). Its vertex is at the origin, \((0,0)\), and it passes through the point \((-3,5)\). It is symmetric about the y-axis.
Key Concepts
Vertex Form of a ParabolaY-Axis Symmetry in ParabolasSketching Parabolas: Practical Tips
Vertex Form of a Parabola
The vertex form of a parabola is a specific way of writing the quadratic equation that emphasizes the vertex's position. When the vertex of the parabola is at the origin and the axis of symmetry is aligned with the y-axis, the vertex form simplifies to:
The variable \(a\) in the equation \(y = ax^2\) controls the parabola's direction and "width".
- \( y = ax^2 \)
The variable \(a\) in the equation \(y = ax^2\) controls the parabola's direction and "width".
- If \(a > 0\), the parabola opens upwards.
- If \(a < 0\), the parabola opens downwards.
Y-Axis Symmetry in Parabolas
Parabolas like the one in this exercise can exhibit symmetry about the y-axis. This symmetry means that if you were to fold the graph of the parabola along the y-axis, both halves would match perfectly.
- This is because, for any point \((x, y)\) on the parabola, there will be a corresponding point \((-x, y)\) that lies directly opposite across the y-axis.
Sketching Parabolas: Practical Tips
Sketching parabolas involves plotting a few key points and understanding the general shape dictated by the parabola's equation. Here's a simple approach:Identify the Vertex:
- For the parabola \( y = \frac{5}{9}x^2 \), the vertex is at the origin, \((0,0)\).
- Since \( a = \frac{5}{9} \) and \(a > 0\), the parabola opens upwards.
- Use symmetry and the equation to find other key points. For instance, if it passes through \((-3, 5)\), it will also pass through \((3, 5)\) due to symmetry.
- Start at the vertex, then draw a smooth curve through the plotted points, ensuring it's symmetrical about the y-axis.
Other exercises in this chapter
Problem 18
Sketch the graph of the given equation. $$ 4(x+3)=(y+2)^{2} $$
View solution Problem 18
Sketch the region that is inside the circle \(r=3 \sin \theta\) and outside the cardioid \(r=1+\sin \theta\), and find its area.
View solution Problem 19
In Problems 17-22, find the Cartesian equations of the graphs of the given polar equations. \(r \cos \theta+3=0\)
View solution Problem 19
In each of Problems 1-20, a parametric representation of a curve is given. (a) Graph the curve. (b) Is the curve closed? Is it simple? (c) Obtain the Cartesian
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