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Eliminate the parameter and obtain the standard form of the rectangular equation. Hyperbola: $x=h+a sec heta, quad y=k+b an heta$...

Question

Eliminate the parameter and obtain the standard form of the rectangular equation. Hyperbola: $x=h+a sec heta, quad y=k+b an heta$

Eliminate the parameter and obtain the standard form of the rectangular equation. Hyperbola: $x=h+a sec heta, quad y=k+b an heta$



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In Exercises 29-32, eliminate the parameter and obtain the standard form of the rectangular equation.

Circle: $\quad x=h+r\ \cos\ \theta, \quad y=k + r\ \sin\ \theta$

Okay, so we have a falling on White. Let's party writing our equation in terms of coastline a bit up inside of it. So we're gonna subject next eight jumbo size and in the Bible, I say we get the people to co sign there. We'll do the same for wife. That's why I came over. He is fine. And if we scrambled right now, get the polling. And we know that identity, that course in core data to find out which one do we know that X minus eight squared over a squared plus wine when it okay squared over peace would be a good one. There's this term right here. Code fine square there. And the term right here. It's fine right there, which is also equal to one.

All right, so when this Question one, eliminate the perimeter and write the rectangular equation. We're gonna use the fact that tan squared data plus one equals Seek it squared beta. Okay, So what do we want to do here? We're going to write secret data in terms of extent that in terms of why, so we get that sequence. Data equals X minus. H over a 10th data equals y minus k. Overby all we did waas subtract caged, caged by a derby. That's it. Okay. And then, by the same logic, we wanted to just square so we can get we could use this equation. So seeking squared data is going to be X minus age squared over a squared tan squared data is going to be why, minus k squared over b squared. And I was just talking to our equation. So tans worth data plus one, we're gonna get Why, minus case weird over B squared plus one equals secret squared data. So ex minus h squared over. He's good. And there we go. We have a rectangular equation.

We are given parametric equations and were asked to eliminate the parameter and obtain the standard form of the rectangular equation. These are for an ellipse. The Parametric equations are X equals H plus a cosign data and why is equal to K plus B sign of data ordinarily, to eliminate a parameter while we need to figure out what parameter we want to eliminate in this case, recall that the standard form of an equation for an ellipse does not include data. Let's try to eliminate data now, as I was saying earlier, ordinarily, he would want to solve one of the equations for data and then plug this into the other equations. However, we have trig functions in this case, so it makes more sense to try to use a trick identity like Pythagorean theorem solving the first equation for co sign data, you get X minus H over A equals co signed data and solving the second equation Y minus K over B equals the sign of data, and therefore we have the X minus h over a squared plus why minus k over B squared equals co sign square data plus sine square data and by Pythagorean theorem. This is one. So we obtain the rectangular equation X minus H squared over a squared plus Y minus K squared over B squared equals one, and this is in fact, the standard form of the rectangular equation or an ellipse.

You're given parametric equations and we are asked to eliminate the parameter contain the standard form of the rectangular equation. These parametric equations are for any hyperbole. To they are X equals H plus A seeking to data and why it was K plus be tangent of data. Now recall that in the standard form of the rectangular equation of a hyper Bella, there is no variable data, so we want to eliminate the parameter data. Ordinarily, what we do is solve one of these equations, Martha, and then plug this into the other equations. However, these equations involved trig functions that makes more sense to try to use a trigger instead. So they solved first equation for seeking to data. I get X minus h over a equals. Seeking to data solving the second equation for tangent data y minus k over B equals the tangent of theta. Now we need to think about what is the relationship between seeking data and tangent data. Well, we know by Pythagorean theorem sine squared data plus coincides square data equals one. If we divide both sides by cosine squared data, we get Hanjin squared of fada plus one equals one over cosine squared data or seek and squared of data. So in this way, we have a relationship. Mm. Secret data in tangent data. Therefore, X minus H over a squared minus y minus k over B squared. This is he can square data minus tangent, squared data. And we see from our derivation from the Pythagorean theorem that this is equal to one. So we've eliminated the parameter data, and so are rectangular equation is X minus H weird ovaries Weird minus y minus k squared over B squared equals one. And this is, in fact, the standard form of the rectangular equation for a hyperbole.


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