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Chapter 11: Problem 19

Exercises \(17-24\) give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch. $$ 2 x^{2}+y^{2}=2 $$

Short Answer

Expert verified
The ellipse is in standard form: \(\frac{x^2}{1} + \frac{y^2}{2} = 1\), with foci at \((0, \pm 1)\).

Step by step solution

01

Write the Equation in Standard Form

The given equation of the ellipse is \(2x^2 + y^2 = 2\). To write it in standard form, we need to have the right-hand side equal to 1. To do this, divide the entire equation by 2. This gives us: \[ \frac{x^2}{1} + \frac{y^2}{2} = 1 \] This equation is now in the form of an ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
02

Identify Ellipse Parameters

From the standard form equation \(\frac{x^2}{1} + \frac{y^2}{2} = 1\), identify the values: - \(a^2 = 1\), so \(a = 1\) - \(b^2 = 2\), so \(b = \sqrt{2}\) Here, since \(b > a\), the ellipse is vertical.
03

Find the Foci of the Ellipse

For an ellipse where \(b > a\), the distance to the foci, \(c\), is found using \(c = \sqrt{b^2 - a^2}\). With \(a = 1\) and \(b = \sqrt{2}\), we have: \[ c = \sqrt{2 - 1} = 1 \] So, the foci are located at \((0, \pm 1)\).
04

Sketch the Ellipse

Draw the ellipse centered at the origin \((0,0)\) with a vertical major axis. - The major semi-axis length \(b = \sqrt{2}\), so it extends \(\pm \sqrt{2}\) units above and below the origin on the y-axis.- The minor semi-axis length \(a = 1\), so it extends \(\pm 1\) units left and right on the x-axis.- Mark the foci at \((0, 1)\) and \((0, -1)\).

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Key Concepts

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

Standard Form of Ellipse
To express an ellipse equation in its standard form, it must fit the template \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This template is pivotal as it allows you to directly identify the properties of the ellipse, such as the major and minor axes. Begin by ensuring the right side of your equation equals 1. In our exercise, the original equation is \( 2x^2 + y^2 = 2 \). To transform it, divide every term by 2, resulting in \( \frac{x^2}{1} + \frac{y^2}{2} = 1 \). This now follows the standard form. Here, the coefficients underneath \(x^2\) and \(y^2\) are effectively \(a^2\) and \(b^2\) respectively.
Ellipse Parameters
Identifying the parameters of an ellipse involves determining the lengths of its axes based on the standard form equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). From this format, the parameters, \(a\) and \(b\), inform the size of the ellipse's axes.
  • Here, \(a^2 = 1\), thus \(a = 1\).
  • For \(b^2 = 2\), \(b = \sqrt{2}\).
As the larger value \(b\) is associated with the \(y\)-term, this ellipse is vertical, signifying a longer vertical axis compared to the horizontal one. These parameters are crucial as they dictate the overall shape and orientation of the ellipse.
Foci of an Ellipse
The foci of an ellipse are two special points along the major axis. They play a key role in defining the geometric properties of an ellipse. Determining their position involves using the calculated values for \(a\) and \(b\). When \(b > a\) in a vertical ellipse, use the formula for the foci distance, \(c = \sqrt{b^2 - a^2}\).
  • From the earlier calculation, \(a = 1\) and \(b = \sqrt{2}\).
  • Calculate \(c\) as \(\sqrt{2 - 1} = 1\).
Thus, the foci are located at \((0, \pm 1)\). These points are symmetric about the center of the ellipse at the origin \((0, 0)\). Understanding the location of foci is important as they highlight the unique property of ellipses: the sum of distances from any point on the ellipse to the two foci is constant.
Graphing Ellipses
Graphing an ellipse involves plotting its shape based on its parameters and foci. Begin by marking the center of the ellipse, which in this case is at the origin \((0, 0)\). From there, use the lengths \(a\) and \(b\) to determine the extremities of the ellipse.
  • The major semi-axis length, \(b = \sqrt{2}\), defines the vertical stretch, extending \(\pm \sqrt{2}\) units from the center on the \(y\)-axis.
  • The minor semi-axis length, \(a = 1\), describes the horizontal reach, stretching \(\pm 1\) unit on the \(x\)-axis.
Identify and plot the foci at \((0, 1)\) and \((0, -1)\), enhancing the visualization of the ellipse's geometric properties. In combination, these steps provide a complete graph of the ellipse, depicting its full scale and positioning with clarity.

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Most popular questions from this chapter

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=6 y, \quad \text { left } 3, \text { down } 2 $$

The asymptotes of \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) Show that the vertical distance between the line \(y=(b / a) x\) and the upper half of the right-hand branch \(y=(b / a) \sqrt{x^{2}-a^{2}}\) of the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) approaches 0 by showing that $$ \lim _{x \rightarrow \infty}\left(\frac{b}{a} x-\frac{b}{a} \sqrt{x^{2}-a^{2}}\right)=\frac{b}{a} \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}-a^{2}}\right)=0 $$ Similar results hold for the remaining portions of the hyperbola and the lines \(y=\pm(b / a) x .\)

Area Find the dimensions of the rectangle of largest area that can be inscribed in the ellipse \(x^{2}+4 y^{2}=4\) with its sides parallel to the coordinate axes. What is the area of the rectangle?

Suspension bridge cables hang in parabolas The suspension bridge cable shown in the accompanying figure supports a uniform load of \(w\) pounds per horizontal foot. It can be shown that if \(H\) is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation $$ \frac{d y}{d x}=\frac{w}{H} x $$ Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that \(y=0\) when \(x=0\) .

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}+2 y^{2}-2 x-4 y=-1 $$
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