Problem 4
Question
Write a polar equation of a conic that has its focus at the origin and satisfies the given conditions. Hyperbola, eccentricity \(\frac{4}{3},\) directrix \(x=-3\)
Step-by-Step Solution
Verified Answer
The polar equation of the hyperbola is \( r = \frac{12}{3 - 4\cos\theta} \).
1Step 1: Understand the Polar Equation of a Conic
The polar equation of a conic with the focus at the origin is given by: \[ r = \frac{ed}{1 - e\cos\theta} \] where \( e \) is the eccentricity, \( d \) is the directrix, and \( \theta \) is the angle in polar coordinates.
2Step 2: Substitute Eccentricity and Directrix Values
For this hyperbola, the eccentricity \( e \) is \( \frac{4}{3} \), and the directrix is perpendicular to the polar axis at \( x = -3 \). Thus, we substitute these values into the equation:\[ r = \frac{\frac{4}{3} \times 3}{1 - \frac{4}{3}\cos\theta} \]
3Step 3: Simplify the Expression
Simplify the numerator:\[ r = \frac{4}{1 - \frac{4}{3}\cos\theta} \]To eliminate the fraction in the denominator, multiply numerator and denominator by 3:\[ r = \frac{12}{3 - 4\cos\theta} \]
4Step 4: Final Polar Equation of the Hyperbola
The polar equation of the hyperbola is:\[ r = \frac{12}{3 - 4\cos\theta} \]This equation describes the conic section with the given eccentricity and directrix.
Key Concepts
HyperbolaEccentricityDirectrix
Hyperbola
To understand the polar equation involving a hyperbola, it's crucial to grasp what a hyperbola is.
A hyperbola is a type of conic section formed when a plane intersects both halves of a double cone.
The shape consists of two symmetrical open curves known as "branches". These branches mirror each other about a central point. In a hyperbola, the distances from any point on the curve to two fixed points (called foci) have a constant difference. This is a unique property that distinguishes hyperbolas from ellipses.
A hyperbola is a type of conic section formed when a plane intersects both halves of a double cone.
The shape consists of two symmetrical open curves known as "branches". These branches mirror each other about a central point. In a hyperbola, the distances from any point on the curve to two fixed points (called foci) have a constant difference. This is a unique property that distinguishes hyperbolas from ellipses.
- The basic form of a hyperbola can be represented as a polar equation, where one focus of the hyperbola is placed at the origin.
- This makes it easier to represent mathematically because the geometry is centered around this fixed reference point.
Eccentricity
Eccentricity is a measure of how much a conic section deviates from being circular.
In simpler terms, it tells us how stretched out a conic section is compared to a circle. For hyperbolas, the eccentricity (\( e \)) is greater than 1.
This means that a hyperbola is more elongated compared to circles or ellipses.
In simpler terms, it tells us how stretched out a conic section is compared to a circle. For hyperbolas, the eccentricity (\( e \)) is greater than 1.
This means that a hyperbola is more elongated compared to circles or ellipses.
- The eccentricity in our hyperbola example is \( \frac{4}{3} \) which is greater than 1.
- This indicates that the shape is wider than a circle.
- The further away the eccentricity value is from 1, the more elongated the shape of the hyperbola becomes.
Directrix
The directrix is an essential line in the construction of conic sections. It works in tandem with the focus to define the shape of the conic.
The directrix is a fixed line, and each conic section maintains a special relationship with its directrix.
The directrix is a fixed line, and each conic section maintains a special relationship with its directrix.
- In the context of the hyperbola provided, the directrix is given as \( x = -3 \)
- This means it's a vertical line that the hyperbola "responds" to in its geometry.
- Every point on the hyperbola has a constant ratio of distances to the focus and to the directrix, determined by the eccentricity.
Other exercises in this chapter
Problem 3
The graphs of \(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1\) and \(\frac{(x-3)^{2}}{5^{2}}+\frac{(y-1)^{2}}{4^{2}}=1\) are given. Label the vertices and foci on
View solution Problem 3
The graph of the equation \(y^{2}=4 p x\) is a parabola with focus \(F(_________ , ____________ )\) and directrix \(x=\) __________________ . So the graph of \(
View solution Problem 4
Determine the \(X Y\)-coordinates of the given point if the coordinate axes are rotated through the indicated angle. $$(-2,1), \quad \phi=30^{\circ}$$
View solution Problem 4
The graphs of \(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1\) and \(\frac{(x-3)^{2}}{4^{2}}-\frac{(y-1)^{2}}{3^{2}}=1\) are given. Label the vertices, foci, and a
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