Problem 25
Question
Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Foci \(F(0, \pm 5), \quad\) conjugate axis of length 4
Step-by-Step Solution
Verified Answer
The equation of the hyperbola is \(\frac{y^2}{21} - \frac{x^2}{4} = 1\).
1Step 1: Understand the Structure of a Hyperbola
The standard form of a hyperbola centered at the origin is either \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), depending on the transverse axis direction. Since the foci are along the y-axis (F(0, \pm c)) with c=5, this hyperbola opens vertically.
2Step 2: Define Variables Based on Given Conditions
For a vertical hyperbola, the relationship between a, b, and c is \(c^2 = a^2 + b^2\). Given that \(c = 5\), we have \(c^2 = 25\). We also know the length of the conjugate axis is 4, which means \(2b = 4\) so \(b = 2\).
3Step 3: Solve for 'a'
Using the relationship \(c^2 = a^2 + b^2\), substitute \(c = 5\) and \(b = 2\) to solve for \(a^2\).\[25 = a^2 + 2^2\]\[25 = a^2 + 4\]\[a^2 = 21\]
4Step 4: Write the Equation of the Hyperbola
Now that we have \(a^2 = 21\) and \(b^2 = 4\), substitute these values into the equation for a vertically oriented hyperbola:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]So the equation is:\[\frac{y^2}{21} - \frac{x^2}{4} = 1\]
Key Concepts
Foci of a HyperbolaConjugate AxisStandard Form of HyperbolaVertically Oriented Hyperbola
Foci of a Hyperbola
In a hyperbola, the foci play a crucial role in its geometric properties. They are two fixed points located symmetrically along one axis of the hyperbola, which is called the transverse axis. For hyperbolas, the equation that relates the foci is expressed as \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to either focus. In this exercise, the foci are at \(F(0, \pm 5)\), indicating that they lie on the y-axis.
This means that the hyperbola is oriented vertically. By knowing the positions of the foci, we can determine the distance \(c\) between the center and a focus - here it's \(c = 5\). Hence, the foci represent points where the difference in distances to any point on the hyperbola remains constant.
This means that the hyperbola is oriented vertically. By knowing the positions of the foci, we can determine the distance \(c\) between the center and a focus - here it's \(c = 5\). Hence, the foci represent points where the difference in distances to any point on the hyperbola remains constant.
Conjugate Axis
The conjugate axis of a hyperbola is perpendicular to the transverse axis. It is not associated with the foci but rather perpendicular to them. The length of the conjugate axis helps to define the shape of the hyperbola.
- In a vertically oriented hyperbola, the conjugate axis runs along the x-axis.
- The conjugate axis has a total length of \(2b\).
- Given the length is 4, we have \(2b = 4\), leading to \(b = 2\).
Standard Form of Hyperbola
The standard form of the hyperbola is instrumental in writing its equation accurately. Depending on the orientation of the hyperbola, we have two possible equations:
This specific structure allows us to explore various properties such as eccentricity, asymptotes, and the overall shape of the hyperbola.
- If the hyperbola opens horizontally, it takes the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
- If it opens vertically, the form changes to \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
This specific structure allows us to explore various properties such as eccentricity, asymptotes, and the overall shape of the hyperbola.
Vertically Oriented Hyperbola
When a hyperbola is vertically oriented, its main characteristics are determined by how it opens and the axes it aligns with. In this orientation, the hyperbola opens upwards and downwards, and its transverse axis is aligned with the y-axis.
This means:
This means:
- The foci are on the vertical or y-axis.
- The standard form shifts to \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
- The vertices, which are closest points on the hyperbola to the center, are located along the y-axis as well.
Other exercises in this chapter
Problem 24
Find an equation of the parabola that satisfies the given conditions. Focus \(F(-3,-2),\) directrix \(y=1\)
View solution Problem 25
Find a polar equation of the conic with focus at the pole that has the given eccentricity and equation of directrix. $$e=\frac{1}{3}, \quad r=2 \sec \theta$$
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Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$2 y=-x$$
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Exer \(19-36:\) Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Vertices \(V(0, \pm 6), \quad\) passing t
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