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It’s your turn on ellipses

solutions to the exercises
11.2
Hello. Welcome to the section, “It’s Your Turn On Ellipses”. In this exercise, we have to compute the center, the foci, and the length of the two axes of the ellipse, which has this question. Good. First of all, let us recall which is the equation of an ellipse with axes which are parallel to the coordinate axis. The equation in its canonical form. OK. The equation is the following. You have x minus x0 squared over a squared plus y minus y0 squared over b squared equal to 1, where a and b are positive real numbers. OK. Which is the meaning of x0, y0, a, and b? OK. The center of this ellipse is the point of coordinates x0, y0. OK.
84.6
Then, the distance between the foci and the center is given by the square root of the module of a squared minus b squared. Good. And finally, 2a and 2b are the lengths of the two axes. Good. Therefore, to solve our exercise, first of all, we have to transform our equation in something which has this shape. Good. Oh, finally, the last thing. Remember that if this denominator is greater than this denominator, then we will have an ellipse which is horizontally oriented, with the major axis, which is parallel to the x-axis. On the other side if this denominator is greater than this one, then we will have an ellipse, which is vertically oriented with the major axis parallel to the y-axis. OK.
163.6
Now let us start with our exercise. Good. We can re-write this equation in the following form. We have x squared minus 4 times 6 plus 4, which multiplies y squared plus a 6 times y plus 24 equal to 0. And now let us complete the squares. And then, these two terms here are equal to what? To x minus 2 squared minus 4 plus 4, which multiplies. OK, now, these two terms here are equal to y plus 3 squared minus 9. And then we have our constant term, 24. Good.
206.9
And now, do you see it’s x minus 2 squared plus 4 times y plus 3 squared and then you have a minus 4 minus 36, which is minus 40, plus 24, which is minus 16. And then moving the minus 16 on the other side with respect to the equal, I have 16. Good. But 16, we know that is 4 squared. And now we divide by 4 squared on both sides. And what do we get? We get x minus 2 squared over 4 squared plus– here we have a 4 here, therefore, dividing it by 4 squared, we have at the denominator we have a 4, which is 2 squared. OK? And then equal to 1. OK.
279.8
We have obtained the equation of our ellipse in this form. And then now we can answer to all the questions of the exercise. Precisely, this is the graph of our ellipse. And now what we can say? We can immediately say which are the coordinates of the center because the center is the point of coordinates 2 and minus 3. Then, as it’s clear from the graph, you see– because this denominator is greater than this denominator, our ellipse is horizontally oriented, and therefore, the foci are on the horizontal line passing through the center C. Good. And what do we know, we know that the distance between the foci and the center is equal to, what?
349.9
To the square root of the module of the difference between these two. But this is 16, this is 4. Then it’s the square root of 12. OK. 12 is 4 times 3. Therefore, we can rewrite this in a simpler way, as 2 times the square root of 3. Good. And knowing that the foci are on the horizontal line passing through the center, and that the distance between the foci and C, and the centre C is 2 times the square root of 3, we immediately get that the focus F1 here is the point– OK.
404.2
Because they are on the same horizontal line as the centre, then the focus F1 will have the second coordinate equal to minus 3 and the first coordinate is 2 minus 2 times the square root of 3. Such a way, the distance is 2 times the square root of 3. And the other focus is F2 equal to 2 plus 2 times, now, the square root of 3 minus 3. Finally. OK. We have that– you know– a here, which is 4 here, is the length of the major axis. The major axis, its length is 8. And the minor axis, has length 4. Good. This is the major axis and this is, here, the minor axis. OK. Good. Thank you very much.

Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.

Exercise 1.

Compute the center, the foci, and the length of the two axes of the ellipse of equation [x^2+4y^2-4x+24y+24=0.] Is it horizontally or vertically oriented?

Exercise 2.

Compute the center, the foci, and the length of the two axes of the ellipse of equation [9x^2+4y^2+18x-24y+9=0.] Is it horizontally or vertically oriented?

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Advanced Precalculus: Geometry, Trigonometry and Exponentials

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