Problem 75 - Calculus Single Variable

Question

Plot the locus of points \(P\) with an ordinate that is equal to the distance of the point \(P\) to the point (0,1)

Step-by-Step Solution

Answer

The locus is the parabola \(y = \frac{x^2 + 1}{2}\).

Step 1: Understanding the Ordinate

The ordinate of a point \(P(x, y)\) is its \(y\)-coordinate. In this problem, the ordinate \(y\) must be equal to the distance \(d\) from \(P\) to the point \((0, 1)\).

Step 2: Calculating the Distance

The distance from a point \((x, y)\) to the point \((0, 1)\) is given by the distance formula: \ \[ d = \sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{x^2 + (y - 1)^2} \]

Step 3: Setting Up the Locus Equation

According to the problem, the ordinate \(y\) is equal to the distance \(d\). Therefore, we set \ \[ y = \sqrt{x^2 + (y - 1)^2} \]

Step 4: Squaring Both Sides

Square both sides of the equation to eliminate the square root: \ \[ y^2 = x^2 + (y - 1)^2 \]

Step 5: Simplifying the Equation

Expand \((y-1)^2\) and reorganize terms: \ \[ y^2 = x^2 + y^2 - 2y + 1 \] \ Cancel \(y^2\) from both sides: \ \[ 0 = x^2 - 2y + 1 \]

Step 6: Solving for y

Rearrange the equation to solve for \(y\): \ \[ 2y = x^2 + 1 \] \ \[ y = \frac{x^2 + 1}{2} \] \ This equation represents the locus of points \(P\).

Key Concepts

Ordinate

In coordinate geometry, the ordinate of a point refers to its vertical position on the Cartesian plane, specifically the y-coordinate. So, when you see a point written as \( P(x, y) \), the term "ordinate" is simply a fancy way of referring to the \( y \) value of that point.
Understanding the ordinate is crucial in solving problems like the one in our exercise, where it is given that the ordinate equals the distance from the point \( P \) to another fixed point.

Distance Formula

The distance formula is a fundamental tool in geometry that calculates the distance between two points in the Cartesian plane. This formula is derived from the Pythagorean theorem and is expressed as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In this problem, we apply the distance formula to find the distance from point \( P(x, y) \) to the fixed point \( (0, 1) \):
* Replace \( x_1 \) with 0 and \( y_1 \) with 1, resulting in:
* \[ d = \sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{x^2 + (y - 1)^2} \]
This calculation helps us express the distance in terms of \( x \) and \( y \), setting the stage for further solving.

Locus Equation

The locus of points refers to the set of all points that satisfy a given condition or equation. In our exercise, the condition is that the ordinate is equal to the distance to a specific point, and our goal is to find this locus equation.
To form this equation, we use the information given: the ordinate \( y \) is equal to the distance \( d \). From earlier calculations, we have:

• Distance: \( \sqrt{x^2 + (y-1)^2} \)
• Ordinate: \( y \)

Equating them gives us: \( y = \sqrt{x^2 + (y-1)^2} \). Square both sides to simplify: \( y^2 = x^2 + (y-1)^2 \). Simplifying further will give us the complete locus equation.

Parabola

While exploring the concept of locus, one common curve encountered is the parabola. A parabola is a U-shaped graph that can open upwards, downwards, or sideways depending on its equation.
In our exercise, finding the locus of points where the ordinate equals the distance leads us to a specific type of parabola, as seen in the final equation:
\[ y = \frac{x^2 + 1}{2} \]
This equation is a typical representation of a parabola opening upwards. Parabolas are an important concept in mathematics as they not only appear in geometric problems but also in real-world applications like physics and engineering. Their unique reflective properties make them useful in satellite dishes and car headlights. Understanding how parabolas form from equations helps demystify their behavior and applications.