4. Graph the conic with focus at the origin, eccentricity 2 and directrix y = 2. Find its polar and rectangular form and show all work to find vertices, foci and asymptotes (if they exist). Be sure to indicate them on the graph.
4. Graph the conic with focus at the origin, eccentricity 2 and directrix y = 2. Find its polar and rectangular form and show all work to find vertices, foci and asymptotes (if they exist). Be sure to indicate them on the graph.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement:
4. Graph the conic with focus at the origin, eccentricity 2 and directrix y = 2. Find its polar and rectangular form and show all work to find vertices, foci and asymptotes (if they exist). Be sure to indicate them on the graph.
### Solution:
To solve this problem, follow these steps:
1. **Identify the Conic Type**:
- Given: Eccentricity \( e = 2 \)
- Directrix: \( y = 2 \)
- Focus at the origin
Since \( e > 1 \), the conic is a hyperbola.
2. **Polar Form**:
- The general polar form of a conic section is:
\[
r = \frac{ed}{1 + e \sin \theta}
\]
- Here, \( e = 2 \) and \( d = 2 \).
Therefore:
\[
r = \frac{2 \cdot 2}{1 + 2 \sin \theta} = \frac{4}{1 + 2 \sin \theta}
\]
3. **Rectangular Form**:
- To convert the given conic section to the rectangular form, we use the relationship between polar and rectangular coordinates:
\[
r = \sqrt{x^2 + y^2}, \quad \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}
\]
- Substituting the above into the polar form:
\[
\sqrt{x^2 + y^2} = \frac{4}{1 + 2 \left(\frac{y}{\sqrt{x^2 + y^2}}\right)}
\]
- Simplify and solve for \( y \):
\[
\sqrt{x^2 + y^2} \left(1 + 2 \frac{y}{\sqrt{x^2 + y^2}}\right) = 4
\]
\[
\sqrt{x^2 + y^2} + 2y = 4
\]
\[
x^2 + y^2 + 2y \sqrt{x^2 + y^2} = 16
\]
- This equation represents the rectangular form of our hyperbola.
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f80a36-5866-477b-b28d-154d74fc6779%2Fb5d6e66f-260d-48f7-8da6-c855dbaf1c96%2F0gayout_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
4. Graph the conic with focus at the origin, eccentricity 2 and directrix y = 2. Find its polar and rectangular form and show all work to find vertices, foci and asymptotes (if they exist). Be sure to indicate them on the graph.
### Solution:
To solve this problem, follow these steps:
1. **Identify the Conic Type**:
- Given: Eccentricity \( e = 2 \)
- Directrix: \( y = 2 \)
- Focus at the origin
Since \( e > 1 \), the conic is a hyperbola.
2. **Polar Form**:
- The general polar form of a conic section is:
\[
r = \frac{ed}{1 + e \sin \theta}
\]
- Here, \( e = 2 \) and \( d = 2 \).
Therefore:
\[
r = \frac{2 \cdot 2}{1 + 2 \sin \theta} = \frac{4}{1 + 2 \sin \theta}
\]
3. **Rectangular Form**:
- To convert the given conic section to the rectangular form, we use the relationship between polar and rectangular coordinates:
\[
r = \sqrt{x^2 + y^2}, \quad \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}
\]
- Substituting the above into the polar form:
\[
\sqrt{x^2 + y^2} = \frac{4}{1 + 2 \left(\frac{y}{\sqrt{x^2 + y^2}}\right)}
\]
- Simplify and solve for \( y \):
\[
\sqrt{x^2 + y^2} \left(1 + 2 \frac{y}{\sqrt{x^2 + y^2}}\right) = 4
\]
\[
\sqrt{x^2 + y^2} + 2y = 4
\]
\[
x^2 + y^2 + 2y \sqrt{x^2 + y^2} = 16
\]
- This equation represents the rectangular form of our hyperbola.
4
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