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Chapter 11: Problem 73

Find a polar equation for each conic section. Assume one focus is at the origin.

Short Answer

Expert verified
Answer: The polar equations for the conic sections with one focus at the origin are as follows: Ellipse: \(r^2 = \frac{a^2b^2}{a^2 \sin^2{\theta} + b^2 \cos^2{\theta}}\) Parabola: \(r = \frac{4a\cos{\theta}}{\sin^2{\theta}}\) Hyperbola: \(r^2 = \frac{a^2b^2}{a^2\sin^2{\theta} - b^2\cos^2{\theta}}\)

Step by step solution

01

Write the general equation of an ellipse

The general equation of an ellipse is given by: (1) \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
02

Convert the general equation into polar coordinates

To convert the ellipse equation (1) into polar coordinates, we use the following conversion formulas: \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\) Substitute these formulas into the ellipse equation (1): \(\frac{(r\cos{\theta})^2}{a^2} + \frac{(r\sin{\theta})^2}{b^2} = 1\)
03

Simplify the polar equation

Now we can simplify the equation: \(r^2\left(\frac{\cos^2{\theta}}{a^2} + \frac{\sin^2{\theta}}{b^2}\right) = 1\) Thus, the polar equation of the ellipse is: \(r^2 = \frac{a^2b^2}{a^2 \sin^2{\theta} + b^2 \cos^2{\theta}}\) #Parabola#
04

Write the general equation of a parabola

The general equation of a parabola is given by: (2) \(y^2 = 4ax\)
05

Convert the general equation into polar coordinates

To convert the parabola equation (2) into polar coordinates, we use the following conversion formulas: \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\) Substitute these formulas into the parabola equation (2): \((r\sin{\theta})^2 = 4a(r\cos{\theta})\)
06

Simplify the polar equation

Now we can simplify the equation: \(r^2\sin^2{\theta} = 4ar\cos{\theta}\) Thus, the polar equation of the parabola is: \(r = \frac{4a\cos{\theta}}{\sin^2{\theta}}\) #Hyperbola#
07

Write the general equation of a hyperbola

The general equation of a hyperbola is given by: (3) \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
08

Convert the general equation into polar coordinates

To convert the hyperbola equation (3) into polar coordinates, we use the following conversion formulas: \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\) Substitute these formulas into the hyperbola equation (3): \(\frac{(r\cos{\theta})^2}{a^2} - \frac{(r\sin{\theta})^2}{b^2} = 1\)
09

Simplify the polar equation

Now we can simplify the equation: \(r^2\left(\frac{\cos^2{\theta}}{a^2} - \frac{\sin^2{\theta}}{b^2}\right) = 1\) Thus, the polar equation of the hyperbola is: \(r^2 = \frac{a^2b^2}{a^2\sin^2{\theta} - b^2\cos^2{\theta}}\) Now, we have the polar equations for all three conic sections. Ellipse: \(r^2 = \frac{a^2b^2}{a^2 \sin^2{\theta} + b^2 \cos^2{\theta}}\) Parabola: \(r = \frac{4a\cos{\theta}}{\sin^2{\theta}}\) Hyperbola: \(r^2 = \frac{a^2b^2}{a^2\sin^2{\theta} - b^2\cos^2{\theta}}\)

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Most popular questions from this chapter

Sketch the graph of the following hyperbolas. Specify the coordinates of the vertices and foci, and find the equations of the asymptotes. Use a graphing utility to check your work. $$10 x^{2}-7 y^{2}=140$$

Find an equation of the following parabolas, assuming the vertex is at the origin. A parabola with focus at (-4,0)

Suppose that two hyperbolas with eccentricities \(e\) and \(E\) have perpendicular major axes and share a set of asymptotes. Show that \(e^{-2}+E^{-2}=1\)

Give the property that defines all parabolas.

Sketch the three basic conic sections in standard position with vertices and foci on the \(x\) -axis.
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