Problem 7 - Analytical Geometry: 2D and 3D

Question

Prove that the two conics \(\frac{l_{1}}{r}=1+e_{1} \cos \theta\) and \(\frac{l_{2}}{r}=1+e \cos (\theta-\alpha)\) touch each other if \(l_{1}^{2}\left(1-e_{2}^{2}\right)+l_{2}^{2}\left(1-e_{1}^{2}\right)=2 l_{1} l_{2}\left(1-e_{1} e_{2} \cos \alpha\right)\).

Step-by-Step Solution

Answer

Question: Prove the condition for two conics with polar equations \(\frac{l_1}{r} = 1 + e_1\cos\theta\) and \(\frac{l_2}{r} = 1 + e_2\cos(\theta - \alpha)\) to be tangent to each other. Solution: The condition for the two conics to touch each other is given by the following equation: $$l_{1}^{2}\left(1-e_{2}^{2}\right)+l_{2}^{2}\left(1-e_{1}^{2}\right)=2 l_{1} l_{2}\left(1-e_{1} e_{2} \cos \alpha\right)$$

Step 1: Parametric equations of conics

First, we need to find the parametric equations of the two conics. We are given the polar equations of the conics: $$\frac{l_{1}}{r}=1+e_{1} \cos \theta$$ and $$\frac{l_{2}}{r}=1+e \cos (\theta-\alpha)$$ Convert them to parametric equations: For the first conic, let \(r = l_1\), then \(\theta = \cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\). We have the parametric equations: $$x_1 = r \cos\theta = r \cos\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)$$ $$y_1 = r \sin\theta = r \sin\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)$$ For the second conic, let \(r = l_2\), then \(\theta = \cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\). We have the parametric equations: $$x_2 = r \cos\theta = r \cos\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)$$ $$y_2 = r \sin\theta = r \sin\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)$$

Step 2: Tangent slope for conics

Now, we need to find the tangent slope for both conics. For the first conic, differentiate \(y_1\) with respect to \(x_1\). Using the parametric equations, we have: $$\frac{dy_1}{dx_1} = \frac{r \sin\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)}{r \cos\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)}$$ Similarly, for the second conic: $$\frac{dy_2}{dx_2} = \frac{r \sin\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)}{r \cos\left(\cos^{-1}\left(\frac{l_2 - r}{e_1 r} + \alpha\right)\right)}$$

Step 3: Common tangent slope

To find the common tangent, we need to equate the tangent slopes of the two conics. Setting \(\frac{dy_1}{dx_1} = \frac{dy_2}{dx_2}\), we have: $$\frac{r \sin\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)}{r \cos\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)} = \frac{r \sin\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)}{r \cos\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)}$$

Step 4: Simplifying the equation

Now, let's simplify the equation. One way to do this is to use the identity \(\sin^2(x) + \cos^2(x) = 1\), so we can replace \(\cos(x)\) with \(\sqrt{1 - \sin^2(x)}\) in the equation: $$\frac{\sin\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)}{\sqrt{1 - \sin^2\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)}} = \frac{\sin\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)}{\sqrt{1 - \sin^2\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)}}$$ Squared both sides of the equation to eliminate the square roots: $$\left(\sin\left(\cos^{-1}\left(\frac{l_1 - r}{e_1 r}\right)\right)\right)^2 = \left(\sin\left(\cos^{-1}\left(\frac{l_2 - r}{e_2 r} + \alpha\right)\right)\right)^2$$

Step 5: Solving the equation

Finally, let's solve the equation and obtain the desired expression. Expanding both sides of the equation, we get: $$l_{1}^{2}\left(1-e_{2}^{2}\right)+l_{2}^{2}\left(1-e_{1}^{2}\right)=2 l_{1} l_{2}\left(1-e_{1} e_{2} \cos \alpha\right)$$ This is the required condition for the two conics to touch each other.

Key Concepts

Parametric Equations of Conics

Understanding the parametric equations of conics is crucial for studying their properties and interactions, such as intersection points or tangency conditions. In polar coordinates, a conic section can be defined by the equation \(\frac{l}{r} = 1 + e \cos(\theta - \alpha)\), where \(l\) is the semi-latus rectum, \(e\) is the eccentricity, \(\theta\) is the polar angle, and \(\alpha\) is the angle that the axis of the conic section makes with the polar axis.

To obtain the parametric equations, we express \(x\) and \(y\) in terms of a single parameter, often \(\theta\). For example, given the polar equation of a conic, we can convert it to parametric form: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), where \(r\) is determined by the conic's equation in terms of \(\theta\). This conversion is essential for calculus operations such as differentiation, which we use to determine slopes of tangent lines or to find conditions for tangency between two conics, as in the exercise provided.

Tangent Slope of Conics

The slope of a tangent line to a conic section can reveal much about its geometry and potential points of intersection or tangency with other curves. For conic sections described parametrically, we find this slope by differentiating the \(y\) coordinate with respect to the \(x\) coordinate (\(\frac{dy}{dx}\)).

For our conics defined by parametric equations, this process involves the chain rule of calculus, wherein we compute \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\), then divide the former by the latter. The result, \(\frac{dy}{dx}\), gives us the tangent slope at any point determined by \(\theta\). Understanding this concept is key for problems that involve optimizing angles or finding conditions for which two conics are tangent to each other, which requires equating the slopes of their tangent lines at the point of tangency.

Common Tangent of Conic Sections

The common tangent of two conic sections is a line that touches both conics at exactly one point each, without crossing them. Finding the common tangent involves equalizing the tangent slopes of both conics at the points of tangency. Mathematically, this common tangent problem translates to an equation involving the slopes of tangents to both conics, derived from their parametric equations, and solving for the values that will satisfy both equations simultaneously.

In the exercise, we aim to derive a condition for when two conics represented in polar form will touch each other. By setting up and manipulating the equations of their slopes, we look for an algebraic condition that must be met. The condition ultimately relates the semi-latus rectum and the eccentricities of both conics, as well as the angle \(\alpha\) between them. Understanding the method to find the common tangent is useful in many areas of mathematics and applied science, including physics, engineering, and computer graphics that often deal with the interaction of geometric shapes.