61–64. Polar equations for conic sections Graph the following ...

Question

61–64. Polar equations for conic sections Graph the following conic sections, labeling vertices, foci, directrices, and asymptotes (if they exist). Give the eccentricity of the curve. Use a graphing utility to check your work.

r = 3/(1 - 2 cos θ)

r = 3/(1 - 2 cos θ)

Accepted Answer

Sketch the graph of R equals 5 divided by 1 minus 2 sine theta. Now, let's look at the standard form for a conic section. This is given by R. It was Edie. Divided by 1 plus or minus e cosine theta. Or ED divided by 1 plus or minus E sine theta. Now let's compare this with our equation, 5 divided by 1 minus 2 sine theta. This follows our second form. So, 5 divided by 1 minus 2 sine theta. This gives us E equals 2. Now, because E equals 2, and this is greater than 1, we have a hyperbole. So with the hyperbola, we first need to find our directors and vertices. So, Our directrix. Which we will put over here. Is given by 2D equals 5. D ends up being 5 halves. This means our directrix is given by Y equals D or Y equals -5 halves. Now we can find our vertices. Let's go ahead and substitute two values. We have our vertices. With theta equals pi divided by 2, and theta equals 3 pi divided by 2. With that, we end up getting R equals 5 divided by 1 minus 2, sin of pi divided by 2. Which just gives us -5. For 3 pi divided by 2, we have 5 divided by 1 minus 2 sin 3 pi divided by 2. Which gives us 5/3 So we now have our polar coordinates, and we can rewrite these in Cartesian coordinates. We can write these as 0, -5. And 0 -5/3. Now we just need the center. Our foresight and our other directors. So then our center. We can find By taking X being equal to 0. This means Y is the average of the Y coordinates of the vertices. So, we have 0. something. Y will be the average of -5 and 5/3. So we have -5 minus 5/3 divided by 2. This gives us a value. Of -10/3, so that is our other value for our center. Now we can find our photo site. And we know that C is the distance between the center and the fossa. And the center is located exactly between the photo site. So, we can say C, first of all. Is 0 minus -1033, which is just 10/33. Which means our second focus is on The opposite side of our first focus, we have YC minus C, or -10/3 minus 10/3, which gives us -203. We now have R2ho sign. We have 00. And 0-203. Finally, we just need our 2nd directrix. Now for our second directrix, we know there's a symmetric about the center. Our first one was a -5 halves. That means our 2nd 1. Will be the distance from the center. Minus our other directrics, so we have 10/3 minus -5 halves with an absolute value. So, our distance then ends up simplifying to give us 5/6, and so our second direct trix is given by Y2. Equals Our center -56. This is -10/3-56, which simplifies to -25/6. We now have everything we need to graph our hyperbole. We'll go up to our graph And first plot the sinner. Machine that was at 0, -103. It will note that as the sinner. If vertices as 0. -5/3. And 0-5. Now we have our directors. Which we know are -5 halves. On our way. And -256. On our way. We'll plot this one. Just below -4. And finally, We can put Our focus Which one of them is at 00. And the other one is at 0-23. Now that we have all of our information, we can draw our two lines for the hyperbola. We'll draw a curve that goes through the vertex and curves around the focus. Both on the top And on the bottom. This ends up being the answer to our problem. OK, I hope to help you solve it. Thank you for watching. Goodbye.

61–64. Polar equations for conic sections Graph the following conic sections, labeling vertices, foci, directrices, and asymptotes (if they exist). Give the eccentricity of the curve. Use a graphing utility to check your work.r = 3/(1 - 2 cos θ)

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61–64. Polar equations for conic sections Graph the following conic sections, labeling vertices, foci, directrices, and asymptotes (if they exist). Give the eccentricity of the curve. Use a graphing utility to check your work.

r = 3/(1 - 2 cos θ)