Problem 10 - Analytical Geometry: 2D and 3D

Question

Find the locus of the foot of the perpendicular from the origin upon the line joining the points \((a \cos \theta, b \sin \theta)\) and \((-a \sin \theta, b \cos \theta)\) where \(a\) is a variable.

Step-by-Step Solution

Answer

Question: Find the equation representing the locus of the foot of the perpendicular from the origin upon the line joining the points (a cos θ, b sin θ) and (-a sin θ, b cos θ) where a is a variable. Solution: 1. Find the equation of the line joining the points P and Q: y - (b sin θ) = m(x - (a cos θ)) 2. Find the equation of the line perpendicular to PQ and passing through the origin: y = -(\(\frac{(-a\sin\theta)-(a\cos\theta)}{(b\cos\theta)-(b\sin\theta)})\) x 3. Find the coordinates of the foot of the perpendicular by solving the two equations: (x, y) = (f(a, b, θ), g(a, b, θ)) 4. Express the coordinates of the foot of the perpendicular in terms of a, b, and θ: x = f(a, b, θ) and y = g(a, b, θ) 5. Find the equation representing the locus of the foot of the perpendicular: h(x, y) = 0

Step 1: Find the equation of the line joining the points P and Q

The given points are P(a cos θ, b sin θ) and Q(-a sin θ, b cos θ). Let's denote the coordinates of P as (x1, y1) and the coordinates of Q as (x2, y2). Now, we can find the slope (m) of the line PQ using the formula: m = \(\frac{y2-y1}{x2-x1}\) Substitute the coordinates of P and Q: m = \(\frac{(b\cos\theta)-(b\sin\theta)}{(-a\sin\theta)-(a\cos\theta)}\) Now, we can find the equation of the line PQ using the point-slope form: y - y1 = m(x - x1)

Step 2: Find the equation of the line perpendicular to PQ and passing through the origin

The slope of the line perpendicular to PQ is the negative reciprocal of m. Let's denote that as m': m' = \(-\frac{1}{m}\) The equation of the line perpendicular to PQ and passing through the origin (0, 0) is given by: y = m'x

Step 3: Find the coordinates of the foot of the perpendicular

We can find the coordinates of the foot of the perpendicular (let's denote it as R) by solving the equations of the lines PQ and the perpendicular equation from Step 2. First simplify both equations: y - (b sin θ) = m(x - (a cos θ)) y = -(\(\frac{1}{m}\))x Now, substitute the value of m from Step 1 into the second equation: y = - \((\frac{(-a\sin\theta)-(a\cos\theta)}{(b\cos\theta)-(b\sin\theta)})\) x Next, solve these two equations to find the coordinates (x, y) of the foot of the perpendicular. The coordinates will be in terms of a, b, and θ.

Step 4: Express the coordinates of the foot of the perpendicular in terms of a, b, and θ

After solving the equations from Step 3, the coordinates (x, y) of the foot of the perpendicular will be in the form: x = f(a, b, θ) y = g(a, b, θ) where f(a, b, θ) and g(a, b, θ) are functions of a, b, and θ.

Step 5: Find the equation representing the locus of the foot of the perpendicular

Use the coordinates (x, y) of the foot of the perpendicular found in Step 4 to find the equation representing the locus of the foot of the perpendicular. Eliminate θ from the coordinates to express the locus in terms of x and y only. The equation will be in the form: h(x, y) = 0 where h(x, y) is a function of x and y. This equation represents the locus of the foot of the perpendicular from the origin upon the line joining the points (a cos θ, b sin θ) and (-a sin θ, b cos θ) where a is a variable.

Key Concepts

Locus in Geometry

The term "locus" in geometry describes a set of points that satisfy a particular condition or rule. In simpler terms, it's the path traced by a point that moves according to a specific condition.
A familiar example is the circle, which is the locus of all points equidistant from a center point. Solving a problem involving the locus requires finding a mathematical expression or equation that defines all the possible positions of a point under given conditions.
In the exercise, we were tasked with finding the locus of the foot of the perpendicular from the origin to a line segment. This involves determining the path formed by this footpoint as the variable, in this case, the angle \(\theta\), alters. This requires drawing on knowledge of lines, slopes, and how changes in the equation impact the path traced by the foot of the perpendicular.

Understanding Perpendiculars

A perpendicular line is simply one that intersects another line at a right angle (90 degrees). This is a crucial concept in geometry and is particularly useful in problems involving right angles or orthogonality.
In coordinate geometry, identifying a perpendicular relationship helps us determine equations of lines and understand spatial relationships between shapes. To find a perpendicular line to a given line, you use the negative reciprocal of the original line’s slope.
For instance, if the slope of a line is \(m\), then the slope of a line perpendicular to it would be \(-\frac{1}{m}\). This switching of slopes is essential for solving step-by-step problems involving perpendicularly in coordinate geometry, as we did in the exercise.

Introduction to Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry to solve problems involving points, lines, and shapes using coordinates. It provides a way to describe the positions of points, lines, and shapes in a plane using numbers.
By using coordinates, we can apply algebra to geometric problems, making it easier to handle and solve them analytically. In coordinate geometry, points are represented as pairs of numbers \((x, y)\) that denote their position on the Cartesian plane.
In our specific exercise, coordinate geometry was employed to find the line joining two points and the locus of another point relative to this line. By representing each point with coordinates \((a \cos \theta, b \sin \theta)\) and \((-a \sin \theta, b \cos \theta)\), we apply algebraic techniques to derive the equations needed to solve the problem effectively.

Equation of a Line

The equation of a line is a crucial tool in both algebra and geometry. It represents a linear relationship between two variables, typically \(x\) and \(y\).
The most common forms of a line equation are the slope-intercept form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept, and the point-slope form \(y - y_1 = m(x - x_1)\), used when you know a point on the line and the slope.
In our problem, we found the equation of the line connecting the points \(P\) and \(Q\) using the point-slope form because we knew coordinates for these points and could calculate the slope. Once the equation is established, it allows us to find other properties like intersection points with other lines or the perpendicular foot drop, tasks necessary for finding the locus of a point.