Problem 35 - Analytical Geometry: 2D and 3D

Question

Find the locus of a point which moves such that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other sides.

Step-by-Step Solution

Answer

Based on the given information, the locus of a point P(x, y) that satisfies the mentioned condition is represented by the equation y^2 = x^2 - h^2, which is a hyperbola with vertices at A(-a, 0) and B(a, 0) and foci at C(0, h).

Step 1: Place the isosceles triangle in a coordinate plane

Place the isosceles triangle in the coordinate plane such that the base lies on the x-axis with equal distances from the origin, meaning that the vertices are A(-a,0), B(a,0), and C(0,h) where a and h are positive constants.

Step 2: Let P(x,y) be the moving point

Let P(x,y) be the point whose locus we need to find. The square of the distance from the base AB of the triangle is given by y^2.

Step 3: Find the distances from the other sides

The distance from point P to sides AC and BC are given by: $$d_{AC}(P) = \frac{|x - h|}{\sqrt{1^2 + (-1)^2}} = |x - h|$$ $$d_{BC}(P) = \frac{|x + h|}{\sqrt{1^2 + (-1)^2}} = |x + h|$$

Step 4: Find the rectangle under distances from sides

The rectangle formed by the distances from sides AC and BC is given by: $$R = d_{AC}(P) \cdot d_{BC}(P) = |x - h||x + h|$$

Step 5: Set up the condition for the locus

We are given that the square of the distance from the base is equal to the rectangle under the distances from the other sides, which can be written as: $$y^2 = |x - h||x + h|$$

Step 6: Simplify the equation

We can simplify the equation by removing the absolute value signs, as y^2 must be non-negative: $$y^2 = (x^2 - h^2)$$

Step 7: Conclusion

The locus of the point P(x,y) is given by the equation: $$y^2 = x^2 - h^2$$ This equation represents a hyperbola with vertices at A(-a,0) and B(a,0) and foci at C(0,h).

Key Concepts

Isosceles Triangle

An isosceles triangle is a special type of triangle that has at least two sides of equal length. This unique characteristic makes it especially interesting in geometry. In an isosceles triangle:

• The two equal sides are known as the 'legs', and the third side is referred to as the 'base'.
• The angles opposite the equal sides are also equal, which is a result of its symmetric properties.
• This symmetry often helps simplify calculations and derivations involving isosceles triangles.

In the given exercise, the isosceles triangle is arranged such that its base lies on the x-axis of a coordinate plane, providing a straightforward setup for further calculations. The base having equal distances from the origin symbolizes the triangle’s symmetric nature on the coordinate plane, making it easier to handle algebraic and geometric operations. Understanding these qualities of the isosceles triangle is crucial when deducing the locus of a moving point relative to it.

Coordinate Plane

The coordinate plane is a fundamental concept in geometry that provides a two-dimensional space for plotting points, lines, and shapes using a pair of numerical coordinates. These coordinates are usually represented as \((x, y)\), where:

• \(x\) is the horizontal position from the origin.
• \(y\) is the vertical position from the origin.

For an isosceles triangle, positioning its base on the x-axis can simplify the problem. The vertices of the triangle can be conveniently labeled as \(A(-a, 0)\), \(B(a, 0)\), and \(C(0, h)\). Here, \(a\) and \(h\) are constants representing specific distances, and this layout allows for easier application of distance formulas and deriving equations. By placing geometric figures on the coordinate plane, we can exploit algebraic methods for solving spatial problems, making abstract ideas more tangible.

Hyperbola

A hyperbola is a type of conic section that can be described as an open curve with two branches. It is defined as the locus of points where the difference of the distances to two fixed points (the foci) is constant. In simpler terms:

• It is a smooth, symmetric curve composed of two separate arcs.
• Each arc is called a 'branch', and they face away from each other.

In the context of the exercise, the equation \(y^2 = x^2 - h^2\) represents a hyperbola. The vertices at \((\pm a, 0)\) and the focus at \((0, h)\) relate to the points of the isosceles triangle placed on the coordinate plane. When solving the exercise, recognizing this equation as a hyperbola helps visualize and understand the locus of points forming this geometric shape in relation to the triangle.

Distance Formula

The distance formula is a crucial tool in geometry for calculating the distance between two points in a coordinate plane. It stems from the Pythagorean theorem and is expressed as:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Here, \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the two points.

In the provided exercise, the distance formula is adapted to find the distance from a point \((x, y)\) to the sides of the triangle. Specifically, it calculates the distance from the point to lines represented by the sides of the isosceles triangle. Then, these distances are used to determine the geometric condition of a locus involving distances. By translating the geometric problem into an algebraic equation, the distance formula becomes an essential method for finding relationships between various distances, further helping to solve for the locus condition given in the problem.