Problem 23
Question
Find the equation of the given central conic. Hyperbola with a vertex at \((0,-4)\) and a focus at \((0,-5)\)
Step-by-Step Solution
Verified Answer
The given data leads to a degenerate case with no valid equation under normal assumptions.
1Step 1: Understand the structure of a hyperbola
A hyperbola with a vertical transverse axis (aligned with the y-axis) can be expressed by the equation \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where \((h, k)\) is the center. In this problem, both the vertex and the focus have x-coordinate 0, so the hyperbola is centered on the y-axis.
2Step 2: Identify the center of the hyperbola
Since the vertex is at \((0, -4)\) and the focus is at \((0, -5)\), we can deduce that the center lies halfway between them. The point halfway between \(-4\) and \(-5\) is \(-\frac{9}{2}\) or \(-4.5\). So the center \((h, k)\) is \((0, -4.5)\).
3Step 3: Calculate the distance from the center to the vertex (a)
The distance from the center \((0, -4.5)\) to the vertex \((0, -4)\) is the value of \(a\). This distance is \(0.5\). This means \(a = 0.5\).
4Step 4: Calculate the distance from the center to the focus (c)
The distance from the center \((0, -4.5)\) to the focus \((0, -5)\) is the value of \(c\). This distance is also \(0.5\), so \(c = 0.5\).
5Step 5: Apply the relationship between a, b, and c
In a hyperbola, the relationship \(c^2 = a^2 + b^2\) holds true. Given that \(a = 0.5\) and \(c = 0.5\), we can solve for \(b\):\[(0.5)^2 = (0.5)^2 + b^2 \]\[0.25 = 0.25 + b^2 \]\[0 = b^2\]Then \(b = 0\).
6Step 6: Write the equation of the hyperbola
Since the value of \(b\) is zero, the hyperbola essentially collapses into a degenerate case. The vertex and the focus coincide with the center, which mathematically implies a contradiction or an ill-defined hyperbola under standard form. Thus, an equation cannot be formed with horizontal/vertical asymmetry parameters.
Key Concepts
Central ConicEllipse and Hyperbola CharacteristicsConic SectionsEquation of a Hyperbola
Central Conic
Central conics form an essential category within conic sections. They refer to ellipses and hyperbolas, both of which have a center. This center is a point equidistant from various points like vertices or foci, specific to the type of conic. In the case of a hyperbola or an ellipse, we find this center by taking the average position between two prominent features, such as vertices or foci, along the axis of symmetry.
- The center of a central conic is a critical part for deriving its equation.
- Understanding how to find the center helps in determining other parameters, such as distances to vertices and foci.
Ellipse and Hyperbola Characteristics
Ellipses and hyperbolas are both conic sections, but they exhibit distinct characteristics that set them apart. They can be defined based on their geometric properties and equations:
- An ellipse is characterized by its two foci and the fact that the sum of distances from any point on the ellipse to these foci is constant.
- A hyperbola, on the other hand, consists of two separate curves whose difference in distances to two foci is constant.
Conic Sections
Conic sections are curves derived from the intersection of a plane and a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. The nature of the section is determined by the angle formed between the plane and the axis of the cone.
Conic sections can be expressed in terms of algebraic equations:
In these equations, the coefficients and terms dictate the specific type and orientation of the conic section.
Conic sections can be expressed in terms of algebraic equations:
- Ellipses and circles have a positive combined constant term in their main equation.
- Hyperbolas, on the other hand, have a difference of squared terms that indicate their nature.
In these equations, the coefficients and terms dictate the specific type and orientation of the conic section.
Equation of a Hyperbola
The equation of a hyperbola is central to identifying its properties and graphically representing it. A hyperbola has an equation of the form:\[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]for a vertical transverse axis, where \(h, k\) is the center.
This highlights the importance of careful attention to the constraints and calculations involved in defining hyperbolas.
- The values \(a\) and \(b\) represent distances to the vertices or the semi-transverse and semi-conjugate axes, respectively.
- In solving for the equation, ensure that you solve for \(b\) using the relationship \(c^2 = a^2 + b^2\).
This highlights the importance of careful attention to the constraints and calculations involved in defining hyperbolas.
Other exercises in this chapter
Problem 23
Name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. $$ r=6 $$
View solution Problem 23
find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=2 \theta^{2}, y=\sqrt{5} \theta^{3} ; \theta \neq 0 $$
View solution Problem 23
Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. $$
View solution Problem 24
Sketch the graph of the given equation. \(25 x^{2}+9 y^{2}+150 x-18 y+9=0\)
View solution