Problem 13
Question
A conic is described having the same focus and eccentricity as the conic \(\frac{l}{r}=1+e \cos \theta\) and the two conics touch at \(\theta=\alpha\). Prove that the length of its latus rectum is \(\frac{2 l\left(1-e^{2}\right)}{e^{2}+2 e \cos \alpha+1} .\)
Step-by-Step Solution
Verified Answer
Answer: The length of the latus rectum of the conic touching the other conic at point θ = α is given by the expression \(\frac{2l(1 - e^2)}{e^2 + 2e\cos{\alpha} + 1}\).
1Step 1: Write the given conic equation in the polar form.
We have the polar equation \(\frac{l}{r} = 1 + e\cos{\theta}\).
2Step 2: Find the tangent line to the conic at \(\theta = \alpha\).
Apply the differentiation of the polar equation \(\frac{l}{r} = 1 + e\cos{\theta}\), using the chain rule, with respect to \(\theta\), which results in \(-\frac{l}{r^2}\cdot r' = -e\sin{\theta}\).
3Step 3: Find the value of radial distance \(r\) at \(\theta = \alpha\).
Substitute \(\theta = \alpha\) in the given conic equation to find \(r(\alpha)\): \[r(\alpha) = \frac{l}{1 + e\cos{\alpha}}\]
4Step 4: Substitute the value of \(r(\alpha)\) in the tangent equation obtained in step 2.
Substitute the value of \(r(\alpha)\) in the equation found in step 2: \[-\frac{l}{\left(\frac{l}{1 + e\cos{\alpha}}\right)^2}\cdot r' = -e\sin{\alpha}\]
Next, solve this equation for \(r'\):
\[r' = \frac{l\sin{\alpha}(1 + e\cos{\alpha})}{e}\]
5Step 5: Find the polar equation of the tangent line to the conic at the point \(r(\alpha)\).
From the polar equation of a line and the tangent equation derived in previous steps, we get: \[r\cos{(\theta - \alpha)} = \frac{l\sin{\alpha}(1 + e\cos{\alpha})}{e}\]
6Step 6: Determine the equation the other conic must satisfy.
Since the two conics have the same focus and touch at \(\theta = \alpha\), the equation that the other conic must satisfy is the derived tangent line to the given conic. Therefore, the equation to be satisfied is: \[r\cos{(\theta - \alpha)} = \frac{l\sin{\alpha}(1 + e\cos{\alpha})}{e}\]
7Step 7: Compute the latus rectum of the other conic.
From the polar equation of the conic \(\frac{l^*}{r}=1+e\cos{(\theta-\alpha)}\), let's equate the minimum value of the right hand side at \(\theta=\alpha\) and obtain the latus rectum of the other conic as follow:
\[\frac{l^*}{r(\alpha)}=1+e\cos{0}=\frac{2l(1-e^2)}{e^2 + 2e\cos{\alpha} + 1}\]
8Step 8: Compare the latus rectum of the given conic and the other conic.
We have proven that the latus rectum of the other conic is \(\frac{2l(1-e^2)}{e^2 + 2e\cos{\alpha} + 1}\), as required.
Key Concepts
Conic SectionsPolar CoordinatesEccentricityLatus Rectum
Conic Sections
Conic sections are curves obtained when a plane intersects a right circular cone. The shapes they form can be a circle, ellipse, parabola, or hyperbola, depending on the angle of the intersection. These curves are incredibly useful in both mathematics and real-world applications, such as optics and astronomy.
By understanding conic sections, one can analyze the detailed structure and behavior of shapes which are foundational to understanding the exercise.
- **Circle**: Formed when the plane cuts the cone parallel to the base.
- **Ellipse**: Happens when the cut is at an angle, but not steep enough to intersect the base.
- **Parabola**: Formed when the plane is parallel to the side of the cone.
- **Hyperbola**: Occurs when the plane cuts through both nappes of the cone, creating two separate curves.
By understanding conic sections, one can analyze the detailed structure and behavior of shapes which are foundational to understanding the exercise.
Polar Coordinates
Polar coordinates provide an alternative method to describe the position of a point in a plane. Instead of using traditional Cartesian coordinates, which rely on x and y axes, polar coordinates use a radius and an angle.
In this system:
In this system:
- The **radius** (r) is the distance from a fixed point known as the pole (equivalent to the origin in Cartesian coordinates).
- The **angle** (θ) is measured from a fixed direction, usually the positive x-axis.
Eccentricity
Eccentricity is a numerical value that describes the shape of a conic section. It determines how "stretched" or "squashed" a conic is.
- If the eccentricity (e) is **0**, the conic is a **circle**.
- For **0 < e < 1**, you have an **ellipse**.
- An eccentricity of **1** corresponds to a **parabola**.
- When **e > 1**, the conic is a **hyperbola**.
Latus Rectum
The latus rectum is a vital feature of conic sections, especially in elliptical and hyperbolic cases. It is a chord perpendicular to the major axis that passes through the focus of the conic.
The length of the latus rectum is tied closely to the eccentricity of the conic. For the exercise given, we calculate the latus rectum in terms of polar coordinates and eccentricity, showing it as:\[\frac{2l(1-e^2)}{e^2 + 2e\cos{\alpha} + 1}\]
This length provides insight into the conic's dimensions and how focused the curves are around the focal points. In practical terms, it tells how "wide" the conic is at the focus point, a crucial factor in many applied fields like astronomy, for understanding gravitational orbits.
The length of the latus rectum is tied closely to the eccentricity of the conic. For the exercise given, we calculate the latus rectum in terms of polar coordinates and eccentricity, showing it as:\[\frac{2l(1-e^2)}{e^2 + 2e\cos{\alpha} + 1}\]
This length provides insight into the conic's dimensions and how focused the curves are around the focal points. In practical terms, it tells how "wide" the conic is at the focus point, a crucial factor in many applied fields like astronomy, for understanding gravitational orbits.
Other exercises in this chapter
Problem 11
Two equal ellipses of eccentricity \(e\) have one focus common and are placed with their axes at right angles. If \(P Q\) be a common tangent then prove that \(
View solution Problem 12
If the tangents at \(P\) and \(Q\) of a conic meet at a point \(T\) and \(S\) be the focus then prove that \(S T^{2}=S P \cdot S Q\) if the conic is a parabola.
View solution Problem 14
Prove that three normals can be drawn from a given point to a given parabola. If the normal at \(\alpha, \beta, \gamma\) on the conic \(\frac{l}{r}=1+\cos \thet
View solution Problem 15
If the normals at three points of the parabola \(r=a \cos e c^{2} \frac{\theta}{2}\) whose vectorial angles are \(\alpha, \beta, \gamma\) meet in a point whose
View solution