The graph of the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) is an ellipse with center (h,k), horizontal axis length 2a, and vertical axis length 2b. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and vertical axis length 2b, and explain why your answer is correct.
I am so, so stuck on this, please help!
they want x and y to be functions of a parameter t, x(t) and y(t) solve the equation for one of the variables, say , y LET x = t y = y(t) you have x and y in terms of t, x(t), y(t)
Good start. To toss out one more idea: What about converting this equation in rectangular coordinates to polar coordinates? Then the parameter coulc be theta, the angle.
\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]
The coding for this is: \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 in case you'd like to compare it to your coding.
yeah the angle could be the parameter trig functions
@mathmale I don't get what you mean. How do you do that? And I don't get how to explain why that would be correct.
@ganeshie8 @freckles I would really appreciate a bit more help if possible, thank you so much!
Not having a textbook on this subject here with me, I did an Internet search for "parametric equations of an ellipse" and quickly found the following: http://www.mathopenref.com/coordparamellipse.html Please take a look at this material. Given the equation of a relationship in cartesian coordinates, you can change that to polar coordinates (which are parametrics) by substituting r cos theta for x and r sin theta for y. Alternatively, you could use the parameter t instead of theta to represent the central angle: substitute r cos t for x and r sin t for y.
Note that the example in http://www.mathopenref.com/coordparamellipse.html features two different magnitudes (a and b) for the parametric substitutions. From the statement of the problem you've posted, you should be able to determine the appropriate coefficients. They're practically given to you.
hint you have \[(\frac{x-h}{a})^2+(\frac{y-k}{b})^2=1 \\ \text{ and you know } \cos^2(t)+\sin^2(t)=1 \]
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