Problem 26
Question
Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the \(x y\) -plane. In each case, find the ellipse's standard-form equation from the given information. $$ \text { Foci: }(0, \pm 4) \quad \text { Vertices: }(0, \pm 5) $$
Step-by-Step Solution
Verified Answer
The standard-form equation of the ellipse is \(\frac{x^2}{9} + \frac{y^2}{25} = 1\).
1Step 1: Understand the Given Information
We have an ellipse centered at the origin with foci located at \((0, \pm 4)\) and vertices located at \((0, \pm 5)\). The foci and vertices along the y-axis indicate a vertical ellipse.
2Step 2: Determine Values of a and c
For vertical ellipses, the vertices are at \((0, \pm a)\). Hence, \(a = 5\) since the vertices are at \((0, \pm 5)\). The foci are given at \((0, \pm c)\), so \(c = 4\) since the foci are at \((0, \pm 4)\).
3Step 3: Use the Relationship to Find b
In an ellipse, the relationship is given by \(c^2 = a^2 - b^2\). Substitute the known values: \(4^2 = 5^2 - b^2\).
4Step 4: Calculate b
Solve the equation \(16 = 25 - b^2\). Rearrange to find \(b^2 = 25 - 16 = 9\), so \(b = 3\).
5Step 5: Write the Standard Form Equation of the Ellipse
The standard form of a vertical ellipse centered at the origin is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\). Substitute the values: \(\frac{x^2}{3^2} + \frac{y^2}{5^2} = 1\). Simplifying, the equation becomes \(\frac{x^2}{9} + \frac{y^2}{25} = 1\).
Key Concepts
FociVerticesEquation of an EllipseEllipse Centered at the Origin
Foci
The foci of an ellipse are like special points. They help define the shape of the ellipse. In our ellipse, the foci are located at
- (0, 4) and (0, -4)
Vertices
Vertices are the points where the ellipse is widest or tallest. In the given problem, the vertices are located at
- (0, 5) and (0, -5)
Equation of an Ellipse
The general formula for an ellipse is a starting point for writing its equation. If we know whether it's more stretched in the x-direction or y-direction, the equation gets simpler. For vertical ellipses like ours, the equation takes the form
- \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \)
- \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \)
Ellipse Centered at the Origin
When we say an ellipse is centered at the origin, it means the middle of the ellipse is at point (0,0) on the coordinate plane. It helps in writing the standard equation without extra terms.
Simply put, there are no shifts needed, either vertically or horizontally. It gives:
Simply put, there are no shifts needed, either vertically or horizontally. It gives:
- The simplest form of the equation
- Ease of calculation and plotting
- \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \)
Other exercises in this chapter
Problem 25
Find the lengths of the curves in Exercises \(21-28 .\) The parabolic segment \(r=6 /(1+\cos \theta), \quad 0 \leq \theta \leq \pi / 2\)
View solution Problem 26
Exercises \(25-28\) give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbol
View solution Problem 26
Find the lengths of the curves. $$ x=t^{3}, \quad y=3 t^{2} / 2, \quad 0 \leq t \leq \sqrt{3} $$
View solution Problem 26
Find a parametrization for the curve. the ray (half line) with initial point \((-1,2)\) that passes through the point \((0,0)\)
View solution