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Chapter 6: Problem 70

Sketch the graph of the ellipse, using latera recta. \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)

Short Answer

Expert verified
The graph of the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\) is an ellipse centered at the origin with vertices at (-2,0) and (2,0), co-vertices at (0,-1) and (0,1), and with latera recta given by the lines \(x = -\sqrt{3} 卤 1/2\) and \(x = \sqrt{3} 卤 1/2\).

Step by step solution

01

Identify the Center

The center of this ellipse is at the origin, (0, 0), since there are no '+ h' or '+ k' in the equation.
02

Identify the Major and Minor Axes

The major axis (a) is the square root of the denominator under \(x^2\), which is \(a=2\). The minor axis (b) is the square root of the denominator under \(y^2\), which is \(b=1\). Therefore, the vertices are at (卤a, 0), which are (-2,0) and (2,0). The co-vertices are at (0, 卤b), which are (0,-1) and (0,1).
03

Identify the Latera Recta

The latera recta are the lines passing through the foci of the ellipse, perpendicular to the major axis, and having length \(2b^2/a\). Since this is an ellipse with major axis along the x-axis, we need to calculate \(b^2/a=1/2\), so the length of the latera recta is 1. They are lines parallel to the y-axis passing through the foci, which are at (卤c,0), where \(c = \sqrt{a^2 - b^2} = \sqrt{3}\). So the equations of the latera recta are \(x = -\sqrt{3} 卤 1/2\) and \(x = \sqrt{3} 卤 1/2\).
04

Sketch the Ellipse

Mark the center, vertices, co-vertices, and latera recta on a graph. Draw the ellipse by making a smooth curve that passes through these points, making it wider along the x-axis. The points where the latera recta intersect the ellipse are also important points to guide the sketching.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Equation
An ellipse is essentially a stretched circle, and its equation is of the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where
  • \(h\) and \(k\) denote the center's x and y coordinates
  • \(a\) and \(b\) represent the horizontal and vertical distances from the center to the ellipse's vertices, respectively
If the equation has no \('+h'\) or \('+k'\), the ellipse is centered at the origin (0, 0). In our exercise, using \(\frac{x^2}{4} + \frac{y^2}{1} = 1\), we can deduce:
  • The center is at (0,0)
  • The horizontal stretch is determined by \(x^2/4\) (making \(a^2 = 4\) or \(a = 2\))
  • The vertical stretch is determined by \(y^2/1\) (making \(b^2 = 1\) or \(b = 1\))
With this simple equation, students can quickly identify these components and understand the ellipse's shape and orientation.
Major and Minor Axes
The major and minor axes are the longest and shortest lines that can be drawn through the center of an ellipse, helping to define its shape. **Major Axis**
  • The major axis runs along the longest path through the center, reaching from vertex to vertex.
  • In our example, where \(\frac{x^2}{4}\) indicates the major axis is on the x-axis, we compute it as \(a = 2\). This axis extends from (-2,0) to (2,0).
**Minor Axis**
  • The minor axis is the shortest path through the center, perpendicular to the major axis.
  • Here, the minor axis is on the y-axis, computed as \(b = 1\), thus running from (0,-1) to (0,1).
The lengths and directions of these axes tell us how elongated the ellipse is and along which axis it stretches the most. Recognizing these axes is crucial in sketching the ellipse accurately.
Foci
The foci of an ellipse are two significant points located along its major axis. They lie inside the ellipse and are key to understanding its geometry.
  • To find the foci, first calculate \(c\) using \(c = \sqrt{a^2 - b^2}\). In our equation, \(a = 2\) and \(b = 1\), so \(c = \sqrt{4 - 1} = \sqrt{3}\).
  • These foci are positioned symmetrically about the center at (卤\(\sqrt{3}\), 0).
The ellipse's definition is based on these points: the sum of the distances from any point on the ellipse to each focus is always constant. This concept helps deeply understand why the ellipse stretches the way it does.
Latera Recta
Latera recta are lines crucial for understanding an ellipse's geometry and assisting in its sketching.
  • They are segments that pass through the foci and are perpendicular to the major axis.
  • Their length provides additional geometric properties, calculated as \(\frac{2b^2}{a}\). In our case, using \(b = 1\) and \(a = 2\), the length of the latera recta is 1.
  • Located at x-coordinates \(\pm \sqrt{3}\), the latera recta are vertical lines passing through points given by \(x = -\sqrt{3} \pm \frac{1}{2}\) and \(x = \sqrt{3} \pm \frac{1}{2}\).
Drawing these lines on a graph assists in visualizing the ellipse, especially in conjunction with the major and minor axes, helping ensure the ellipse correctly captures all significant geometrical features.

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Most popular questions from this chapter

Find the standard form of the equation of the ellipse with the given characteristics. Vertices: (5,0),(5,12)\(;\) endpoints of the minor axis: (1,6),(9,6)

Find the vertex, focus, and directrix of the parabola, and sketch its graph. \(y=\frac{1}{4}\left(x^{2}-2 x+5\right)\)

Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. . \(9 x^{2}+9 y^{2}+18 x-18 y+14=0\)

Find an equation of the tangent line to the parabola at the given point, and find the \(x\) -intercept of the line. \(x^{2}=2 y,(4,8)\)

Find the standard form of the equation of the ellipse with the given characteristics. Vertices: (0,2),(4,2)\(;\) endpoints of the minor axis: (2,3),(2,1)
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