Questions and Answers #4 Conic Sections

Questions and Answers Sheet 4 Conic Sections Question #1 Is 𝑥 2 = 4𝑎𝑦 a function while 𝑦 2 = 4𝑎𝑥 is not? Answer: This is typically written as 𝑓: 𝑋 ⇒ 𝑌 . The symbol that stands for an arbitrary input, or the representing element of the domain, is called the independent variable. And similarly the dependent variable. It is a tradition to write x for the independent variable and y for the dependent variable. Now, first, let me assume 𝑎 ≠ 0 . Otherwise you wouldn't have meaningful functions, but two equations. Also, I will assume that your domain and codomain are the set of real numbers. In your first example 𝑦= 𝑥2 . 4𝑎 Using this formula for each real numbered value of x, you can compute a value for y. Hence, by definition, we have a function. On the other hand, the second formula is 𝑦 2 = 4𝑎𝑥, which produces a quadratic equation for y as soon as we assign a value for x. For example, lets set x=a , then 𝑦 2 = 4𝑎2 which has two solutions 𝑦 = ±2𝑎 . This isn't unique; hence doesn't agree with the definition of a function. Not only that, suppose a is positive, then for x=-1 , we get 𝑦 2 = −4𝑎 which is impossible. This means that there is no value assigned to -1. One can restrict the domain and codomain to non-negative real values to remedy these two problems. Then even the second formula defines an honest function. Question #2 Is there a parametrization of a hyperbola 𝑥 2 − 𝑦 2 = 1 by functions x(t) and y(t) both birational? Consider the hyperbola 𝑥 2 − 𝑦 2 = 1. I am aware of some parametrizations like: 1. (𝑥(𝑡), 𝑦(𝑡)) = ( 2. (𝑥(𝑡), 𝑦(𝑡)) = ( 𝑡 2 +1 𝑡 2 −1 2𝑡 𝑡 2 +1 , , 2𝑡 2𝑡 𝑡 2 −1 𝑡 2 −1 ) ) 3. (𝑥(𝑡), 𝑦(𝑡)) = (cosh 𝑡 , sinh 𝑡) 4. (𝑥(𝑡), 𝑦(𝑡)) = (sec(𝑡) , tan(𝑡)) The first and the second are by rational functions x(t) and y(t). But the functions are not birational(i.e. with rational inverse of each). Is there a parametrization where: - x(t) is rational with inverse also rational, and - y(t) is rational with inverse also rational? Is possible, to find a parametrization where both are rational and at least one of the has inverse rational? Answer: If (f(t),g(t)) is a parameterization with f and g rational and 𝑔−1 is rational, then: (𝑓(𝑔−1 (𝑠)), 𝑠) is a parameterization and 𝑓 ∘ 𝑔−1 is rational But 𝑓 ∘ 𝑔−1 (𝑠) = √1 + 𝑠 2 is not a rational function. This works if we even just want (f(t),g(t)) to parameterize an subset of the curve 𝑥 2 − 𝑦 2 = 1 , for t in some interval (a,b). This same argument shows that f can’t be birational if g is rational. Question #3 Name or Adjective for Ellipse with very different (or very similar) scales I am looking for an adjective or word to describe an ellipse (or ellipsoid, in more dims) where the length of the principal axes are of roughly the same order of magnitude O(a)=O(b) (including equal, i.e. a circle). For instance: - a=2, b=5 would be something like an "isometric" ellipse (or ellipsoid) - a=2, b=50 would be something like an "asometric" ellipse (or ellipsoid) since they are of different orders of magnitude Answer: I am aware of no special name. But is a name was to be given, I would say very plainly "with high or low" eccentricity due to formula 𝑒 = √1 − 𝑏2 on a scale 0 ≤ 𝑒 < 1 𝑎2 (the minimum 0 for a circle, 1 cannot be attained because we would have then a parabola). Besides, for the 3D case of an ellipsoid with semiaxes 𝑎 ≥ 𝑏 ≥ 𝑐, I would like to attract your attention on two specific names for the case of equality: - if a=b>c : "Prolate spheroid" (in fact ellipsoid) (rugby ball), - if a>b=c : "Oblate spheroid" (ideal Cinderella's pumpkin...). The prolate spheroidals, in particular, are connected with various applications: - a 3D coordinates' system - orthogonal functions. See for example this article explaining how that these functions are at the same time orthogonal in 𝐿2 (−1,1) and in 𝐿2 (−∞, +∞) Question #4 (2𝑥 + √4𝑥 2 + 1)(√𝑦 2 + 4 − 2) ≥ 𝑦 > 0 minimum vale of x+y Answer: Explanation: 𝑔(𝑦) = 𝑦 √𝑦 2 +4−2 is monotonic decreasing and 𝑔(𝑦) ≥ 1 for 𝑦 ≥ 0 , and function 𝑓(𝑥) = 2𝑥 + √4𝑥 2 + 1 is monotonic increasing and 𝑓(𝑥) ≥ 1 requires 𝑥 ≥ 0 . First we could easily find out that when x+y is minimized, 𝑓(𝑥) = 𝑔(𝑦) = 𝑧 ≥ 1 so 𝑥 = 𝑓 −1 (𝑧) = So 𝑥 + 𝑦 = 𝑧 2 −1 4𝑧 + 4𝑧 𝑧 2 −1 𝑧 2 −1 4𝑧 𝑎𝑛𝑑 𝑦 = 𝑔−1 (𝑧) = 4𝑧 𝑧 2 −1 . ≥ 2 , where 𝑧 ≥ 1 The minimal value is reached when 𝑧 2 − 1 = 4𝑧. Question #5 Rotation matrix to construct canonical form of a conic 𝐶: 9𝑥 2 + 4𝑥𝑦 + 6𝑦 2 − 10 = 0. I've found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial 2  9 −t p(t ) = det    2 6−t  The eigenvalue are 𝑡1 = 5, 𝑡2 = 10, with associated eigenvectors (-1, 2), (2, 1) . Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that det(𝑅) = 1 ): R= 1  1 2   5  −2 1  Then (𝑥, 𝑦)𝑡 = 𝑅(𝑥 ′ , 𝑦 ′ )𝑡 , and after some computations I find the canonical form 1 ′2 4 ′2 𝑥 + 𝑦 = 1. 2 5 Answer: Explanation: If 𝑞(𝑥, 𝑦) = 9𝑥 2 + 4𝑥𝑦 + 6𝑦 2 − 10 and 𝑥 ′ = 1 √5 (𝑥 + 2𝑦) and 𝑦 ′ = 1 √5 (−2𝑥 + 𝑦), 1 then 𝑞(𝑥 ′ , 𝑦 ′ ) = 5𝑥 2 + 10𝑦 2 − 10. So, 𝑞(𝑥 ′ , 𝑦 ′ ) = 0 ⟺ 𝑥 2 + 𝑦 2 = 1. 2 Question #6 Shape induced by the condition of angle bisector passing through a fixed point Let A, B be two points on 2D plane. For any 𝐶 ∈ 𝐴𝐵 , define the set 𝑆 = {𝑃 | ∠𝐴𝑃𝐶 = ∠𝐶𝑃𝐵 }. What is the shape of S? Answer: Step 1 Here is a solution using coordinate geometry. If |AB| = k and |𝐴𝐶| |𝐵𝐶| = 𝑡 where 𝐶 ∈ 𝐴𝐵 . Using angle bisector theorem, | AP | | AC | = = t | AP |2 = t 2  | PB |2 | PB | | BC | WLOG, coordinates of A is (0,0) and of B is (k,0). If coordinates of P is (x,y), we get ~𝑥 2 + 𝑦 2 = 𝑡 2 ((𝑥 − 𝑘)2 + 𝑦 2 ) Step 2 If t = 1, x = k i.e. the locus is perpendicular bisector of AB. 2 If t  1, x + y − 2 (𝑥 − 𝑘𝑡 2 𝑡 2 −1 2 2 2kt 2 k 2t 2 x = − t 2 −1 t 2 −1 ) + 𝑦2 = ( 𝑘𝑡 𝑡 2 −1 ) 2 which is a circle with center at ( 𝑘𝑡 2 𝑡 2 −1 Question #7 Solve below somewhat symmetric equations: x, y, z subject to 𝑥 2 + 𝑦 2 − 𝑥𝑦 = 3 (𝑥 − 𝑧)2 + (𝑦 − 𝑧)2 − (𝑦 − 𝑧)(𝑥 − 𝑧) = 4 (𝑥 − 𝑧)2 + 𝑦 2 − 𝑦(𝑥 − 𝑧) = 1 𝑥, 𝑦, 𝑧 ∈ 𝑅+ Answer: 𝑘𝑡 , 0) and radius |𝑡 2 −1| . We can take a quadrilateral ABCD, for which: AC  BD = {E}, AC = BD = z, ED = x, EC = y, AEB = 60 ,, AB = 2, BC = 1 and CD = 3 Thus, we obtain your conditions: 𝑥 2 + 𝑦 2 − 𝑥𝑦 = 𝐸𝐷2 + 𝐸𝐶 2 − 2𝐸𝐷 ⋅ 𝐸𝐶 cos 6 0∘ = 𝐶𝐷2 = 3, (𝑥 − 𝑧)2 + (𝑦 − 𝑧)2 − (𝑦 − 𝑧)(𝑥 − 𝑧) = (𝑧 − 𝑥)2 + (𝑧 − 𝑦)2 − (𝑧 − 𝑥)(𝑧 − 𝑦) = = 𝐸𝐵2 + 𝐸𝐴2 − 2𝐸𝐵 ⋅ 𝐸𝐴 cos 6 0∘ = 𝐴𝐵2 = 4 and (𝑥 − 𝑧)2 + 𝑦 2 − 𝑦(𝑥 − 𝑧) = (𝑧 − 𝑥)2 + 𝑦 2 + 𝑦(𝑧 − 𝑥) = = 𝐸𝐵2 + 𝐸𝐶 2 − 2𝐸𝐵 ⋅ 𝐸𝐶 cos 1 20∘ = 1. Now, let BFCD be a parallelogram. Thus, 𝐴𝐶 = 𝑧 = 𝐵𝐷 = 𝐹𝐶 and ACF = DEC = 60 , which gives that △ 𝐴𝐹𝐶 is equilateral triangle, 𝐵𝐴 = 2, 𝐵𝐹 = 𝐶𝐷 = √3 and BC=1 Now, let R be a rotation around C by −60∘ and 𝑅({𝐵}) = {𝐺} thus, since △ 𝐵𝐺𝐶 is an equilateral triangle, we obtain: 𝐵𝐺 = 1, 𝐹𝐺 = 𝑅(𝐴𝐵) = 2 and 2 𝐹𝐺 2 = 22 = (√3) + 12 = 𝐵𝐹 2 + 𝐵𝐶 2 , which says FBG = 90 , FBC = FBC + GBC = 90 + 60 = 150 , which gives 𝑧 2 = 𝐴𝐶 2 = 𝐹𝐶 2 = 𝐹𝐵2 + 𝐵𝐶 2 − 2𝐹𝐵 ⋅ 𝐵𝐶 cos 1 50∘ = 3 + 1 + 3 = 7, whisch gives 𝑧 = √7 and from here we get the answer: (𝑥, 𝑦, 𝑧) = ( 5 , 1 √7 √7 , √7). Question #8 What are the equations of rotated and shifted ellipse, parabola and hyperbola in the general conic sections form? How will look the general conic sections equation 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 in case of rotated and shifted from coordinates origin ellipse, parabola and hyperbola? I need a formulas for coefficients A, B, C, D, E and F for ellipse, hyperbola and parabola. I did it for notrotated conic sections at the origin of coordinates but have a difficults with rotated and shifted. For i.e. standart ellipse equation 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 gives me general equation 1 𝑎2 𝑥 2 + 0𝑥𝑦 + 1 𝑏2 𝑦 2 + 0𝑥 + 0𝑦 − 1=0 I need the same in case if ellipse located in position (h;k) and rotated on some angle 𝛼 from positive X axis. And the same for parabola and hyperbola. Answer: There is a simple way to derive these formulas. Let's take the case of the ellipse. Originally you have 𝑥2 𝑎2 + 𝑦2 𝑏2 =1 If you rotate the coordinate system by an angle 𝜃 , you get 𝑥 → 𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑦 → 𝑥 sin 𝜃 + 𝑦 cos 𝜃 𝑥 → 𝑥 cos 𝜃 − 𝑦 sin 𝜃 𝑦 → 𝑥 sin 𝜃 + 𝑦 cos 𝜃 So the equation becomes (𝑥 cos 𝜃−𝑦 sin 𝜃)2 𝑎2 + (𝑥 sin 𝜃+𝑦 cos 𝜃)2 𝑏2 =1 If you translate the origin by (𝑥0 , 𝑦0 ) , the equation above transforms to (𝑥 cos 𝜃−𝑦 sin 𝜃+𝑥0 )2 𝑎2 + (𝑥 sin 𝜃+𝑦 cos 𝜃+𝑦0 )2 𝑏2 =1 Now all you need to do is to expand the parentheses, and group the terms. Question #9 What is the easier way to find the circle given three points? Given three points (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ), and (𝑥3 , 𝑦3 ) , if 𝑦2 − 𝑦1 𝑦3 − 𝑦2 𝑦1 − 𝑦3 ≠ ≠ , 𝑥2 − 𝑥1 𝑥3 − 𝑥2 𝑥1 − 𝑥3 then there will be a circle passing through them. The general form of the circle is 𝑥 2 + 𝑦 2 + 𝑑𝑥 + 𝑒𝑦 + 𝑓 = 0. By substituting 𝑥 = 𝑥𝑖 and 𝑦 = 𝑦𝑖 , there will be a system of equation in three variables, that is:  x1   x2 x  3 y1 1  d   −( x12 + y12 )      y2 1  e  =  −( x22 + y22 )  . y3 1  f   −( x32 + y32 )  As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error. Is there a better way to solve for the equation of the circle? Answer: Move the circle so that it passes through the origin (subtract a point from all three). The equation loses a coefficient, 𝑥 2 + 𝑦 2 + 𝑑𝑥 + 𝑒𝑦 = 0 which is easier to solve. Then reverse-translate. x12 + y12 y1 2 2 d x + y2 y2 − = 2 x y1 2 2 1 x2 y2 x12 + y12 x22 + y22 e − = x y1 2 2 1 x2 y2 x1 x2 are the coordinates of the center in the moved position.