Problem 19
Question
Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. Vertices: (0,±4) passes through the point (3,1)
Step-by-Step Solution
Verified Answer
The equation of the ellipse in standard form is \(\frac{x^2}{0.6} + \frac{y^2}{16} = 1\).
1Step 1: Identify Values
From the given vertices of the ellipse, 'b' can be determined. Vertices are (0,±4) so 'b' = 4.
2Step 2: Substitute the point into the Ellipse Equation
The equation for an ellipse in standard form is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). We know that 'b=4', but 'a' is still unknown. Substitute point (3,1) into the equation and solve for 'a'. This gives \(\frac{3^2}{a^2} + \frac{1^2}{4^2} = 1\).
3Step 3: Solve for 'a'
On simplifying the above equation, we get \(\frac{9}{a^2} + \frac{1}{16} = 1\). Solving this equation for 'a^2', gives 'a^2' = 9/15 = 0.6
4Step 4: Write the Standard Form of the Ellipse
Now that we have 'a^2' and 'b^2', we can write the standard form of the ellipse, which is \(\frac{x^2}{0.6} + \frac{y^2}{16} = 1\).
Key Concepts
Standard FormVerticesOriginConic Sections
Standard Form
The standard form of the equation for an ellipse gives us a precise mathematical way to describe its shape. It helps in defining the ellipse's attributes such as its axes, vertices, and orientation. The equation of an ellipse in standard form is typically written as:
- For a horizontally oriented ellipse: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
- For a vertically oriented ellipse: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)
Vertices
Vertices are key points on an ellipse that define its shape and size. In a typical ellipse, the vertices lie on the major axis, the longest diameter of the ellipse. These points are located at distances of \( \pm a \) and \( \pm b \) from the center, depending on the orientation of the ellipse. For our given exercise, the vertices are \( (0, \pm 4) \), indicating that the major axis is vertical. This tells us that \( b = 4 \). The vertices are crucial for determining the semi-axes since they provide the range of the ellipse along its major axis.
Origin
The origin \( (0,0) \) serves as a common reference point in coordinate geometry. When an ellipse is centered at the origin, it greatly simplifies the equation.The center \( (h,k) \) effectively becomes \( (0,0) \), making the standard equations of the ellipse easier to manage:
- Without shifts: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Conic Sections
Ellipses are one type of conic section, a curve formed by the intersection of a cone with a plane. These curves include circles, ellipses, parabolas, and hyperbolas.
An ellipse is created when the intersecting plane cuts through a right circular cone at an angle, resulting in a closed curve. Here are a few key points about ellipses as conic sections:
- An ellipse might appear circular but is distinguished by a stretched, oval shape.
- The defining attribute is the consistent sum of distances from any point on the ellipse to two fixed points called foci.
- Ellipses have practical applications in astronomy, orbital mechanics, and various fields of engineering.
Other exercises in this chapter
Problem 19
Write the equation of the circle in standard form. Then identify its center and radius. $$\frac{1}{4} x^{2}+\frac{1}{4} y^{2}=1$$
View solution Problem 19
Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (0,4),(0,0) passes through the point \((\sqrt{5},-1)\)
View solution Problem 20
Identify the type of conic represented by the equation. Use a graphing utility to confirm your result. $$r=\frac{9}{3-2 \cos \theta}$$
View solution Problem 20
Test for symmetry with respect to the line \(\theta=\pi / 2,\) the polar axis, and the pole. $$r=4 \csc \theta \cos \theta$$
View solution