A polar conic section Consider the equation r² = sec2θ
b...

Question

A polar conic section Consider the equation r² = sec2θ

b. Find the vertices, foci, directrices, and eccentricity of the curve."

Accepted Answer

Hello. In this video we are going to be finding the vertices, foci, and eccentricity of the polar curve R squared equal to 2 sequent of 2 theta by converting it into Cartesian coordinates. So in order to approach this problem, let's quickly recall the identities we use to convert between polar and Cartesian coordinates. The first identity is X is equal to our cosine of theta. The second identity will be y is equal to our sin of theta. The third is R2 is equal to x2 + y2, and the fourth identity will be theta is equal to tangent inverse of y divided by x. So we are going to need to use these four identities in order to help us convert the given polar curve into Cartesian coordinates. Now again, the polar curve given to us is R2 equal to 2 multiplied by sequent of 2 theta. Now, the left hand side is easy to rewrite because we just need to use the identity for R2, and the identity for R2 is X2 + Y2. So the left hand side can be rewritten as X2 + Y2. Now the right hand side is going to require a little bit more work to simplify. The first thing we want to do is we want to try to simplify sequent of 2 theta, but recall that the identity for sequent is that the inverse identity is 1 divided by cosine of theta. So the first thing we can do is apply this identity to simplify the right hand side. So we can rewrite the right hand side as 2 multiplied by 1 divided by cosine of 2 theta. Now what we need to do is we need to use a double angle identity to simplify cosine of 2 theta. Well, one of the angle identities that we have is that cosine of 2 theta. is equal to cosine 2 minus sin 2. So applying this identity allows us to rewrite the denominator as 2 multiplied by 1 divided by cosine squared theta minus sine square of theta. Now, the last thing we want to do is we want to try to rewrite cosine squared and sin squared using the conversion identities between polar and Cartesian coordinates, or polar and Cartesian. So, if we take a look at the first identity, we have X. Is equal to r cosine of theta. If we were to solve for cosine in this case, we get cosine theta is equal to x divided by r. And if we take a look at the second identity, y is equal to r sin of theta, and solving for sine of theta in this case leaves us with sine theta is equal toy divided by r. So if we were to apply these identities to cosine squared and sine squared, we can further rewrite the right hand side of the polar curve to be 2 multiplied by 1 divided by the quantity x divided by R2 minus the quantity y divided by R2. We can further simplify this to be 2 divided by X2 divided by R2 minus Y2 divided by R2. Now if we multiplied both the numerator and denominator by R2, then we can further rewrite the right hand side to be 2 R 2. Divided by x2 minus y2. So this is going to be the simplified form. In fact, we can even apply one more identity. If we replace R 2d by using the third identity for the Cartesian conversions, then we can further rewrite the right hand side to be 2 multiplied by the quantity x2 + y2 divided by x2 minus y2. So now that we've simplified the left and right-hand side, let's go ahead and place this back into the polar curve. So, by using these identities, we can rewrite the polar curve to be X2 + Y2 equal to 2 multiplied by the quantity X2 + Y2 divided by X2 minus Y2. So now what we need to do is we need to go ahead and simplify. The first thing we can do is we can go ahead and multiply both sides by x2 minus y2. That will leave us with x2 minus y2 multiplied by the quantity x2 + y2, and this will equal the 2 multiplied by x2 + y2. And since we have X2 + Y2 on both sides of the equation, we can divide these quantities out and simplify the polar curve to be X2 minus Y2 equal to 2. So this is going to be the simplest form of the polar curve in Cartesian coordinates, but keep in mind, in the original problem, we want to find the vertices, the foci, and the eccentricity of the polar curve. So we need to be able to identify what kind of conic we are working with. Since we want to identify the conic, the first thing we want to do is we want to set this equation equal to 1. In order to do this, we will go ahead and divide every term in this equation by 2. That will leave us with x2 divided by 2 minus y2 divided by 2 is equal to 1. Now, what kind of conic are we working with? Well here, this equation matches the form of x2 divided by a 2. Minus Y2 divided by b2 is equal to 1. So what this is, is this is the equation of a hyperbola. And it's the equation of a hyperbola where the major axis is going to be the x-axis. So now that we have identified what kind of conic we are working with, we can go ahead and find the foci eccentricity and vertices. Let's go ahead and start by finding the foci of the conic. Now, in this case here, the identity of the foci of a hyperbola with the major axis being the x axis is defined as positive and negative C0. And this is where we have the identity C2 is equal to A2 + B2. Now if we are comparing the general tonic. To the polar curve that we simplified earlier, this tells us that A2 is equal to 2 and B2 is equal to 2 as well. This also tells us that A is equal to the square root of 2, and B is equal to the square root of 2 as well. So by plugging in the values, we get C2 is equal to A2, which is 2 + B2, which is 2. This will simplify to be c 2 is equal to 4, and by taking the square root we get cs equal to positive and -2. So what this tells us is that the foci of the conic is going to be defined as two points, positive and negative, 2.0. So this is going to be one of the solutions to the current problem. The next thing we want to find is the eccentricity. Now the eccentricity is defined as e is equal to c divided by a. Now here, if we're going back to the foci, we know that the value of C is going to be positive and negative too, but in this case here we will work with the positive quantity, and we know the value of A is square root of 2. So if we were to plug these values into the eccentricity, the eccentricity will be defined as 2 divided by the square root of 2. And by rationalizing the denominator, we get an eccentricity of square root of 2. This is going to be the 2nd solution to this problem. Finally, we need to find the vertices. Now the vertices for a hyperbola with the major axis being the x axis is defined as positive and negative a 0. Now earlier we did define the value of A. A is equal to the square root of 2. So if we were to plug this into the identity of the vertices, there would be two vertices, and the identity of the vertices is going to be positive and negative square root of 20. This will be the final solution to the problem. So just to recap, we were able to use the conversion identities to rewrite the given polar curve into the rectangular curve X2 divided by 2 minus Y2 divided by 2 is equal to 1. This matched the identity of a hyperbola with the major axis being the x axis, and the foci is positive and -20. The vertices are positive and square root 2.0, and the eccentricity is the square root of 2. With that being said, the overall solution to this problem is going to bed. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

A polar conic section Consider the equation r² = sec2θb. Find the vertices, foci, directrices, and eccentricity of the curve."

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A polar conic section Consider the equation r² = sec2θ
b. Find the vertices, foci, directrices, and eccentricity of the curve."