Problem 22 - Analytical Geometry: 2D and 3D

Question

Prove that the exterior angle between any two tangents to a parabola is equal to half the difference of the vectorial angles of their points of contact.

Step-by-Step Solution

Answer

Question: Prove that the exterior angle between any two tangents to a parabola is equal to half the difference of the vectorial angles of their points of contact. Solution: To prove this, we first find the equation of the tangent lines at the points of contact on the parabola. We then calculate the exterior angle between these tangent lines using the difference of their slopes. Finally, we show that this exterior angle is equal to half the difference of the vectorial angles of the points of contact, completing the proof.

Step 1: Find the equation of the tangent lines

Let's assume that we have a parabola with equation \(y = ax^2 + bx + c\), and assume that two points on the parabola, \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\), are the points of contact for the two tangent lines. Since these points are on the parabola, we have \(y_1 = ax_1^2 + bx_1 + c\) and \(y_2 = ax_2^2 + bx_2 + c\). The slope of a secant line from point \(P_1\) to point \(P_2\) can be found as \(m_{12} = \frac{y_2 - y_1}{x_2 - x_1}\). Now we can find the derivative of the parabolic function to find the slope of the tangent lines: $$\frac{dy}{dx} = 2ax + b$$ We can then find the slopes of the tangent lines at points \(P_1\) and \(P_2\) as: $$m_{1} = 2a x_1 + b$$ $$m_{2} = 2a x_2 + b$$

Step 2: Calculate the exterior angle between the tangent lines

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be found by: $$\tan(\theta) = \frac{m_2 - m_1}{1 + m_1 m_2}$$ Since we want the exterior angle between the tangent lines, we will consider the supplementary angle, \(180^\circ - \theta\), or in radians, \(\pi - \theta\). This can be found by taking the negation of the previous result, denoted as \(\theta_{ext}\): $$\tan(\theta_{ext}) = \frac{m_1 - m_2}{1 + m_1 m_2}$$

Step 3: Show that the exterior angle is equal to half the difference of the vectorial angles of their points of contact

The vectorial angles of the points of contact are essentially just the slope of the tangent lines in polar coordinates. Thus, we have: $$\phi_1 = \arctan(m_1)$$ $$\phi_2 = \arctan(m_2)$$ The difference between these angles, divided by 2, is: $$\frac{\phi_2 - \phi_1}{2} =\frac{\arctan(m_2) - \arctan(m_1)}{2}$$ Using the tangent addition formula, we can relate the difference of the arctangent functions to the tangent of the exterior angle: $$\tan\left(\frac{\phi_2 - \phi_1}{2}\right) = \frac{\tan(\phi_2) - \tan(\phi_1)}{1 + \tan(\phi_1) \tan(\phi_2)} = \frac{m_2 - m_1}{1 + m_1 m_2}$$ This result is the same as the result for the tangent of the exterior angle between the two tangent lines, \(\tan(\theta_{ext})\). Thus, we have proven that the exterior angle between any two tangents to a parabola is equal to half the difference of the vectorial angles of their points of contact.

Key Concepts

Tangent lines

Tangent lines are crucial concepts in mathematics, especially when dealing with curved shapes like parabolas. Imagine a car driving along a winding path; at any point, the direction the car is pointed can be thought of as a tangent line to the curve. For a parabola, a tangent line will touch the curve at exactly one point—this point is known as the point of contact.

To find this tangent line, we calculate the slope at the point of contact. This is done using the derivative of the function that describes the parabola. The slope tells us how steep the tangent is at that exact location.

• The equation of the parabola is often written as: \[ y = ax^2 + bx + c \]
• The derivative, giving us the slope of the tangent line, is: \[ \frac{dy}{dx} = 2ax + b \]

These lines are especially important because they provide a linear approximation of the curve at the point of contact.

Exterior angle

The exterior angle is the angle between two lines—here, it's the angle formed by two tangents meeting at a point outside of the parabola. Imagine two lines each representing a different road. The angle between these roads, as they meet at a junction, is similar to the exterior angle between two tangents meeting outside the parabola.

For two lines with slopes \(m_1\) and \(m_2\), the angle between them can be found using:

• \( \tan(\theta) = \frac{m_2 - m_1}{1 + m_1 m_2} \)

Since we are interested in the exterior angle, we might actually be considering its supplementary, i.e., the angle that makes 180 degrees with \(\theta\):

• \( \tan(\theta_{ext}) = \frac{m_1 - m_2}{1 + m_1 m_2} \)

This external perspective helps understand how two tangents diverge from one another as they extend away from their points on the parabola.

Vectorial angles

Vectorial angles are a clever way to describe directions. Imagine standing at the top of a hill and spinning around; every direction you face can be translated into an angle. For tangents to a parabola, this concept helps describe the direction of each tangent line at its point of contact.

When you look at the slope of a tangent line, it essentially tells you the direction in which the line is heading. By finding the angle that matches this slope, we get the vectorial angle:

• \( \phi_1 = \arctan(m_1) \) and \( \phi_2 = \arctan(m_2) \)

These angles are crucial because they provide a "vectorial" perspective of our tangent directions, letting us compare how two tangents might differ in angle rather than just slope. This comparison is paramount in geometric proofs, as often the difference in such angles tells you more about the shape and behavior of the elements involved.

Derivative of a function

The derivative acts like a mathematical magnifying glass. It allows us to see how a function is behaving at any given point. Specifically, it gives us the slope of the tangent line to the curve of the function—telling us the curve's direction and speed (rate of change) at that point.

For a parabola expressed as \( y = ax^2 + bx + c \), the derivative is:

• \( \frac{dy}{dx} = 2ax + b \)

This formula indicates how the y-value (vertical position) changes with the x-value (horizontal position).

The process of differentiation extracts this slope information, allowing us to define tangent lines precisely. Used broadly in calculus, derivatives help in optimizing functions, understanding motions, and even in predicting rates of change in various fields. The concept is fundamental to solving a wide range of real-world problems, from physics to finance.