Question

A polar conic section Consider the equation r^2 = sec 2 theta. Convert the equation to Cartesian coordinates and identify the curve

Conic sections
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asked 2020-12-30
A polar conic section Consider the equation \(r^2 = sec 2 \theta\). Convert the equation to Cartesian coordinates and identify the curve

Answers (1)

2020-12-31

Given: we have an equation in polar form \(r^2 = \sec 2 \theta\) Part a) to determinate the equation to Cartesian coordinates and identify the curve Explain: to convert the equation we can write the equation sa follows \(r = \sqrt{\sec 2 \theta}\)
\(r = {1}{\sqrt{\cos 2 \theta}}\) We know these relations \(x = r \cos \theta, y = r \sin \theta, r = \sqrt{x^2 + y^2}\) Can be written \(\cos \theta\) \(= x/r, \sin \theta = y/r\) \(r = \frac{1}{\sqrt{\cos^{2} \theta}} = \frac{1}{\sqrt{\cos^2 \theta - \sin^2 \theta}} [\cos^{2} \theta = \cos^2 \theta - \sin^2 \theta]\)

Putting in the equation we have \(r = \frac{1}{\sqrt{\frac{x}{r}}}^2 - \frac{y}{r}^2\) \(\frac{x^2}{r^2} - \frac{y^2}{r^2} = \frac{1}{r^2}\) We have the equation \(x^2 - y^2 = 1\)

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