Problem 38
Question
For the parabola \(y^{2}=4 p x\) in Figure \(12, P\) is any of its points except the vertex, \(P B\) is the normal line at \(P, P A\) is perpendicular to the axis of the parabola, and \(A\) and \(B\) are on the axis. Find \(|A B|\) and note that it is a constant.
Step-by-Step Solution
Verified Answer
The length \(|AB|\) is \(2p\), and it's a constant.
1Step 1: Identify the Equation of the Parabola
The given equation is \(y^2 = 4px\). This is the equation of a parabola oriented horizontally, with its vertex at the origin \((0,0)\) and its focus \((p, 0)\). The directrix is at \(x = -p\).
2Step 2: Determine the Coordinates for P
Let the coordinates of point \(P\) on the parabola be \((x_0, y_0)\). Given that \(P\) lies on the parabola, we have \(y_0^2 = 4px_0\).
3Step 3: Calculate the Equation of Normal Line PB
The slope of the tangent line at \(P\) is \(\frac{dy}{dx} = \frac{2p}{y_0}\). The slope of the normal line is the negative reciprocal, which is \(-\frac{y_0}{2p}\). The equation of the normal line through point \(P\) is \(y - y_0 = -\frac{y_0}{2p}(x - x_0)\).
4Step 4: Find Point B on the x-axis
Point \(B\) is where the normal meets the x-axis. Set \(y = 0\) in the normal equation: \(0 - y_0 = -\frac{y_0}{2p}(x - x_0)\). Solving this gives \(x = x_0 - 2p\). Thus, \(B = (x_0 - 2p, 0)\).
5Step 5: Determine Point A on the y-axis
Since \(PA\) is perpendicular to the axis of the parabola, it is a vertical line through \(P\), thus \(x = x_0\) for point \(A\). On the y-axis, \(A = (x_0, 0)\).
6Step 6: Calculate the Distance |AB|
\(|AB|\) is simply the distance between the x-coordinates of \(A\) and \(B\), which is \(|x_0 - (x_0 - 2p)| = |2p|\).
7Step 7: Conclusion: Verify Constant Distance
Since \(2p\) is a constant value, the distance \(|AB| = 2p\) is always constant for any point \(P\) on the parabola.
Key Concepts
Equation of ParabolaNormal Line CalculationGeometry of ParabolasConstant Distance Property
Equation of Parabola
The equation of the parabola given in this exercise is expressed as \( y^2 = 4px \). Knowing the form of this equation is crucial because it reveals a lot about the parabola's characteristics. This specific form indicates a horizontal orientation.
- **Vertex**: The vertex of this parabola is located at the origin, point \((0,0)\).- **Focus**: The focus is positioned at \((p, 0)\), where \( p \) is a positive constant.- **Directrix**: The directrix of the parabola is the vertical line \( x = -p \).
This equation is a hallmark in the study of conic sections. It signifies that the parabola is situated symmetrically around the x-axis. Such forms simplify many calculations related to the geometry and algebra of the parabola, including lines such as normals and tangents.
- **Vertex**: The vertex of this parabola is located at the origin, point \((0,0)\).- **Focus**: The focus is positioned at \((p, 0)\), where \( p \) is a positive constant.- **Directrix**: The directrix of the parabola is the vertical line \( x = -p \).
This equation is a hallmark in the study of conic sections. It signifies that the parabola is situated symmetrically around the x-axis. Such forms simplify many calculations related to the geometry and algebra of the parabola, including lines such as normals and tangents.
Normal Line Calculation
To understand the relationship between the parabola and its normals, we must delve into calculating the normal line at a given point on the parabola.
Given a point \( P(x_0, y_0) \) on the parabola, the slope of the tangent at \( P \) is found using implicit differentiation as \( \frac{dy}{dx} = \frac{2p}{y_0} \).
The normal line plays a pivotal role in connecting the geometry of the parabola to points on the x-axis, such as point \( B \). Calculating normals efficiently can illustrate concepts such as symmetry and constant distance properties.
Given a point \( P(x_0, y_0) \) on the parabola, the slope of the tangent at \( P \) is found using implicit differentiation as \( \frac{dy}{dx} = \frac{2p}{y_0} \).
- Tangent Slope: \( \frac{2p}{y_0} \)
- Normal Slope: The negative reciprocal of the tangent slope is \( -\frac{y_0}{2p} \).
The normal line plays a pivotal role in connecting the geometry of the parabola to points on the x-axis, such as point \( B \). Calculating normals efficiently can illustrate concepts such as symmetry and constant distance properties.
Geometry of Parabolas
The shape and properties of parabolas make them fascinating geometric figures. Here, we explore aspects like symmetry and focus-related properties.
- **Opening Direction**: The orientation of the parabola \( y^2 = 4px \) tells us it opens horizontally.- **Symmetry**: Parabolas are symmetric around their axis. For this specific equation, the axis of symmetry is along the x-axis.- **Vertex and Focus Relationship**: The vertex is the point where the parabola reaches its minimum or maximum width. The focus is a point of convergence for all reflected lines that are parallel to the principal axis of the parabola.
These properties are crucial when considering transformations or scaling of the parabola. Knowing the geometry helps in understanding more advanced topics, such as the effects of different values of \( p \) on the parabola's shape and positioning.
- **Opening Direction**: The orientation of the parabola \( y^2 = 4px \) tells us it opens horizontally.- **Symmetry**: Parabolas are symmetric around their axis. For this specific equation, the axis of symmetry is along the x-axis.- **Vertex and Focus Relationship**: The vertex is the point where the parabola reaches its minimum or maximum width. The focus is a point of convergence for all reflected lines that are parallel to the principal axis of the parabola.
These properties are crucial when considering transformations or scaling of the parabola. Knowing the geometry helps in understanding more advanced topics, such as the effects of different values of \( p \) on the parabola's shape and positioning.
Constant Distance Property
One of the significant properties of a parabola is the constant distance property, which is demonstrated when calculating the distance between points \( A \) and \( B \) on the axis in this problem. As derived:
- For any point \( P \) on the parabola, the distance \( |AB| \) is exactly \( 2p \).
This happens because point \( A \) is vertically aligned with \( P \) and point \( B \) is where the normal line intercepts the x-axis.
- For any point \( P \) on the parabola, the distance \( |AB| \) is exactly \( 2p \).
This happens because point \( A \) is vertically aligned with \( P \) and point \( B \) is where the normal line intercepts the x-axis.
- Coordinate of \( A \): \((x_0, 0)\)
- Coordinate of \( B \): \((x_0 - 2p, 0)\)
Other exercises in this chapter
Problem 38
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