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It’s your turn on level curves of quadratic functions – 1 –

It's your turn on level curves of quadratic functions
Hello, welcome to the section of exercises, “It’s Your Turn on Level Curves of Quadratic Functions.” Good. We wanted to study the curves represented by these three equations. In particular, we want to understand which is the type and to recognize, I mean, the type of conic that you get. Good. I decided to divide this video in three parts, one for each of these three equations. Good. Then let us start with the first one. The strategy, which is general, is the following. First of all, we have to compute the discriminant of our curve. If the discriminant will be negative, then we will be in the elliptic case. If the discriminant is 0, we will be in the parabolic case.
And if the discriminant is positive, we will be in the hyperbolic case. Let us compute the discriminant in our situation. The discriminant delta is the square of the coefficient of the term in x y, which is 0 in this case, minus 4 times the coefficient of x squared, which is 1, times the coefficient of y squared, which is 1 again. And we get minus 4. Therefore, we are in the elliptic case. Good. When you have a curve of elliptic type, then you have the following possibilities. Or you have an ellipse, and in particular, for example, also a circle, or otherwise you have the empty set. How can you distinguish between these two situations? One idea is the following.
You can consider, for example, all the horizontal lines in the plane. And you look for intersections between these horizontal lines and our curve. If you don’t find any intersection for all the resultant lines, then surely the curve represented by our equation will be the empty set. Otherwise we will have an ellipse. Good. Let us see in our situation what we have. We can intersect the generic horizontal line, y equals a, where a is a real number with our equation, x squared plus y squared minus 4 times x plus 6 times y minus 3 equals 0. Attention, now you have not to compute, really, the intersection.
So you have only to understand if there are intersections between horizontal lines and our curve. But in this case, it’s really very easy because if you take a equal to 0, then we have y equals 0. And substituting in this equation we get x squared minus 4 times x minus 3 equal to 0. And now this system has solutions indeed. You take, of course, y equals 0. And you see this is an equation of the second degree. And it has solutions. It has a couple of distinct solutions. Why? Because, for example, if you consider the discriminant of this equation of a second degree, what do you get?
You get 16 minus 4 times the coefficient of x squared times minus 3, which is greater than 0. Therefore you have solutions for this equation. It is not important to compute the solutions but only to know that there are solutions. Then our curve surely is not the empty set. By the way, now, look, what do we have in this curve? There are no terms in x y. The coefficient of x squared is equal to the coefficient of the y squared. Then, therefore, we have a circle.
By the way, let us compute exactly which circle we have. You see, we can rewrite this equation now in the following form, x squared minus 4 times x plus y squared plus 6 times y minus 3 equal to 0. And now we complete the squares. This term here is equal to what? To x minus 2 squared minus 4. Good. And these terms here are equal to what now? To y plus 3 squared minus 9. And then we have minus 3 equal to 0. Therefore we get x minus 2 squared plus y plus 3 squared equal to what? To 16.
Then we have a circle of centre C of coordinates 2, minus 3 and the radius r equal to 4.

Do your best in trying to solve the following problem. In any case it is solved in the video and in the pdf file below.

Exercise 1. (first part)

Say if the curve represented by the following equations are elliptic, parabolic or hyperbolic and recognise the type of conic it describes: [1. x^2+y^2-4x+6y-3=0;]

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Advanced Precalculus: Geometry, Trigonometry and Exponentials

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