Questions and Answers #3 Conic Sections

Questions and Answers Sheet 3 Conic Sections Question #1 Angle between normal vector of ellipse and the major-axis. I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape 𝑥 = 𝑎 cos(𝑡) , 𝑦 = 𝑏 sin(𝑡) with the parameter t reffering to Ellipse in polar coordinates. I need to solve it for any angle t. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be tan(𝜙) = 𝑎 tan(𝑡) Derivation. However this is very similar to the polar angle namely its simply the term a and b 𝑏 flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong? I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of (\phi) then I would indeed get by combing 𝑡 = 𝑎 𝑎 𝑏 𝑏 tan−1 ( tan(𝜃)) and 𝜙 = tan−1 ( tan(𝑡)) 𝑎 𝑎𝑏 tan 𝜃 𝑎2 𝜙 = tan−1 ( tan (tan−1 𝑏𝑖𝑔 ( 𝑔))) = tan−1 ( 2 tan 𝜃) 𝑏 𝑏𝑖 𝑏 My excuse for my rambling, I find these angles confusing... Answer: The tangent vector to the ellipse at parameter value t is (−𝑎 sin 𝑡 , 𝑏 cos 𝑡), so the tangent of the angle \psi it makes with the positive x-axis is given by 𝑏 tan 𝜓 = − cot 𝑡. 𝑎 The angle 𝜙 to the outward normal vector is 𝜓 − 𝜋/2 , so 𝜋 𝑎 2 𝑏 tan 𝜙 = tan (𝜓 − ) = − cot(𝜓) = tan 𝑡, as desired. Question #2 Can it be shown that gives a conic section? 𝑟2 = 1 1−𝜀 cos(2𝜃) Answer: Rewrite it as 𝑟 2 (1 − 𝜖 cos(2𝜃)) = 1 ⟺ 𝑟 2 (1 − 𝜖(cos 2 𝜃 − sin2 𝜃)) = 1 and then expand. We have 𝑟 2 = 𝑥 2 + 𝑦 2 , 𝑟 2 cos 2 𝜃 = 𝑥 2 and 𝑟 2 sin2 𝜃 = 𝑦 2 so we get the traditional equation of an ellipse or a hyperbola. Question #3 Convert hyperbola in rectangular form to polar form 3𝑦 2 − 16𝑦 − 𝑥 2 + 16 = 0. Answer: Staring with 3𝑦 2 − 16𝑦 − 𝑥 2 + 16 = 0 Put it first in the standard format. Compute the square in y 8 2 64 3 3 3 (𝑦 − ) − − 𝑥 2 + 16 = 0 8 2 16 3 3 3 (𝑦 − ) − 𝑥 2 = 9 8 2 3 2 (𝑦 − ) − 𝑥 =1 16 3 16 Thus the center is at 8 4 3 3 (0, ) , 𝑎 = and 𝑏 = 4 √3 The focal distance 1 1 8 9 3 3 𝑐 = √𝑎2 + 𝑏 2 = 4√ + = 16 Hence, the foci are at (0,0) and (0, ) 3 Taking the focus that is at the origin, then 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃. Back to the original equation 3𝑦 2 − 16𝑦 − 𝑥 2 + 16 = 0 Substituting the polar expressions, 3(𝑟 sin 𝜃)2 − 16𝑟 sin 𝜃 − 𝑟 2 cos 2 𝜃 + 16 = 0 Using cos 2 𝜃 = 1 − sin2 𝜃 and collecting terms 𝑟 2 (4 sin2 𝜃 − 1) − 16𝑟 sin 𝜃 + 16 = 0 From the quadratic formula, and taking the positive root 𝑟= 1 2(4 sin2 𝜃−1) (16 sin 𝜃 − √256 sin2 𝜃 − (256 sin2 𝜃 − 64)) And this simplifies to 𝑟= 1 2(4 sin2 𝜃−1) (16 sin 𝜃 − 8) Factoring the denominator and cancelling equal terms 𝑟= 4 2 sin 𝜃+1 Question #4 Coordinate geometry and Trignometry. Find the condition so that the line px +qy=r intersects the ellipse angles differ by 𝜋 4 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 in points whose eccentric . Though I know how to solve it using parametric coordinates, I was wondering if there's an another approach which is less time consuming. Answer: 𝜋 Let 𝑃1 = (𝑎 cos 𝑡 , 𝑏 sin 𝑡) and 𝑃2 = (𝑎 cos(𝑡 + 𝜙) , 𝑏 sin(𝑡 + 𝜙)) where 𝜙 = , then we need to satisfy two 4 conditions: 1. 𝑃1 is on the line px+qy=r , therefore pa cos t + qb sin t = r (1) 2. 𝑃2 is on the line as well, hence, pa cos(t +  ) + qb sin(t +  ) = r (2) This last equation becomes cos t ( pa cos  + qb sin  ) + sin t ( − pa sin  + qb cos  ) = r (3) And after substituting cos 𝜙 = sin 𝜙 = 1 √2 , it becomes, cos t ( pa + qb) + sin t (qb − pa) = 2r (4) Now solving (1) and (4) for cost and sint, we find that cos 𝑡 = −𝑟 sin 𝑡 = −𝑟 (𝑞𝑏−𝑝𝑎)−√2𝑞𝑏 𝑝2 𝑎2 +𝑞 2 𝑏 2 √2𝑝𝑎−(𝑝𝑎+𝑞𝑏) 𝑝2 𝑎2 +𝑞 2 𝑏 2 Since cos 2 𝑡 + sin2 𝑡 = 1, then the condition is 2 (𝑝2 𝑎2 +𝑞 2 𝑏 2 ) 𝑟2 = (𝑞𝑏 − 𝑝𝑎)2 + (𝑝𝑎 + 𝑞𝑏)2 + 2(𝑞 2 𝑏 2 + 𝑝2 𝑎2 ) − 2√2(𝑞𝑏(𝑞𝑏 − 𝑝𝑎) + 𝑝𝑎(𝑝𝑎 + 𝑞𝑏)) Simplifying the right hand side, 2 (𝑝2 𝑎2 +𝑞 2 𝑏 2 ) 𝑟2 = 4(𝑝2 𝑎2 + 𝑞 2 𝑏 2 ) − 2√2(𝑞 2 𝑏 2 + 𝑝2 𝑎2 ) Dividing through by (𝑝2 𝑎2 + 𝑞 2 𝑏 2 ), yields, 𝑝2 𝑎2 +𝑞 2 𝑏 2 𝑟2 = 4 − 2√2 And this is the condition on the line. Question #5 Determine whether (𝑥, 𝑦) = (𝑚 cos(𝜔𝑡) , 𝑛 cos(𝜔𝑡 − 𝜙)) describes an ellipse for all m, 𝑛, 𝜔, 𝜙 Answer: Step 1 𝑥 = 𝑚 cos 𝜔 𝑡 𝑦 = 𝑛 cos(𝜔𝑡 − 𝜙) = 𝑛(cos 𝜔 𝑡 cos 𝜙 + sin 𝜔 𝑡 sin 𝜙) This is a linear system in cos 𝜔 𝑡 and sin 𝜔 𝑡 , the system is given by 0  cos t   x  = n sin    sin t   y   m n cos   The solution (assuming sin 𝜙 ≠ 0 and 𝑛 ≠ 0 ) is cos t   n sin  1  sin t  = mn sin  −n cos     0   x m  y  Step 2 Now, since cos 2 𝜔 𝑡 + sin2 𝜔 𝑡 = 1, then the equation governing x,y is x  n2 y   −nm cos  −nm cos    x  2 2 2    = m n sin  2 m   y written concisely, 𝑟 𝑇 𝑄𝑟 = 𝑐 The principal determinants are 𝑛2 and 𝑛2 𝑚2 (1 − cos 2 𝜙) = 𝑛2 𝑚2 sin2 𝜙 So to have a positive definite matrix Q, we must also have that 𝑚 ≠ 0 If these three conditions are met, then the above equation represents an ellipse, because then the positive definite matrix Q can be diagonalized into 𝑄 = 𝑅𝐷𝑅𝑇 where D is the diagonal matrix of Q's eigenvalues, and R is the rotation matrix whose columns are the corresponding (orthogonal) unit eigenvectors. Now the equation above can be re-written as 𝑟 𝑇 𝑅𝐷𝑅𝑇 𝑟 = 𝑐 Divide through by c, 𝑟 𝑇 𝑅𝐸𝑅𝑇 𝑟 = 1 1 with 𝐸 = 𝐷 𝑐 Now let 𝑣 = 𝑅𝑇 𝑟 so that 𝑟 = 𝑅𝑣 , then 𝑣 𝑇 𝐸𝑣 = 1 and this is clearly an equation of an ellipse because the diagonal entries of E are both positive. Finally the vector r is just a rotation about the origin of vector v, thus it describes a rotated ellipse whose angle of rotation is determined by the angle between the first column of R and the positive x axis. Question #6 Eliminate the parameter t from ℎ= 𝑘= 3𝑡 2 −4𝑡+1 𝑡 2 +1 4−2𝑡 𝑡 2 +1 Answer: The circle 𝑆(𝑥, 𝑦) = 0 passes through A(3, 0), B(1, 4) and C(3, 4). Since ∠𝐴𝐶𝐵 = 90∘ , the center D of the circle is the midpoint of the line segment AB. Let E be the tangent point. Then, note that △ 𝑂𝐷𝐸 is a right triangle where DE is the radius. Question #7 Equation of a section plane in hyperbolic paraboloid Find the equation of a plane passing through Ox and intersecting a hyperbolic paraboloid 𝑥2 𝑝 − 𝑦2 𝑞 = 2𝑧(𝑝 > 0, 𝑞 > 0) along a hyperbola with equal semi-axes. My attempt: The equation of a plane passing through Ox is By+Cz=0. So 𝑧 = −𝐵𝑦 𝐶 into the equation of hyperbolic paraboloid 𝑥2 𝑝 − 𝑦2 𝑞 = −2𝐵𝑦 𝐶 and transform to to do next? I don't understand how to find B and C if we assume 𝑥2 𝑝 equal semi-axes. Answer: A generic plane containing the x axis (Ox) has a unit normal vector 𝑛 = (0, cos 𝜃 , sin 𝜃) − 𝐵𝑞 2 (𝑦− ) 𝐶 𝑞 −𝐵𝑦 𝐶 𝑥2 𝑝 =− . Now substitute 𝑧 = − 𝐵2 𝑞 𝐶2 𝐵𝑞 2 ) 𝐶 (𝑦− 𝑞 =− 𝐵2 𝑞 𝐶2 . What is a hyperbola with Two vectors (unit and mutually perpendicular) that span this plane are 𝑣1 = (1,0,0) and 𝑣2 = (0, sin 𝜃 , − cos 𝜃) Hence, on this plane, the position vector of any point is 𝑟 = [𝑣1 , 𝑣2 ]𝑢 where 𝑢 = [𝑢1 , 𝑢2 ]𝑇 is the coordinate vector of r with respect to 𝑣1 , 𝑣2 . Hence, x = u_1 𝑦 = sin 𝜃 𝑢2 𝑧 = − cos 𝜃 𝑢2 Plug this into the equation of the hyperbolic paraboloid, 𝑢12 sin2 𝜃 𝑢22 − = −2 cos 𝜃 𝑢2 𝑆 𝑝 𝑞 which is an equation of hyperbola in 𝑢1 and 𝑢2 . If the semi-axes are equal then 𝑝= 𝑞 sin2 𝜃 Hence, the required angle 𝜃 satisfies |sin 𝜃| = √ 𝑞 𝑝 This determines four possible values for 𝜃 and correspondingly four planes. Question #8 Find domain and range of the slanted hyperbola Given the conic section 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 and I know that it is a hyperbola and 𝐵 ≠ 0. How to find its domain and range? I guess the method of Lagrange multipliers will fail here. Answer: Explanation: To find a range, fix y and consider a quadratic equation with respect to x: 𝐴𝑥 2 + (𝐵𝑦 + 𝐷)𝑥 + 𝐶 2 𝑦 + 𝐸𝑦 + 𝐹 = 0 If 𝐴 = 0 and 𝐵 = 0, there is 𝑥 = − 𝐶 2 +𝐸𝑦+𝐹 𝐷 for each y and the domain is (−∞, ∞)(𝐷! = 0 , because, otherwise, the conic is degenerated). 𝐷 𝐶 2 +𝐸𝑦+𝐹 𝐵 𝐵𝑦+𝐷 If 𝐴 = 0 and 𝐵 ≠ 0, there is no x for 𝑦 = − , for other values of y it is 𝑥 = − and the domain is 𝐷 (−∞, ∞){− }. 𝐵 If A!=0 , qudratic equation has at least one solution, when its discriminant 𝑑 = (𝐵𝑦 + 𝐷)2 − 4𝐴(𝐶 2 + 𝐸𝑦 + 𝐹) ≥ 0. So, we need to find y, for which (𝐵𝑦 + 𝐷)2 − 4𝐴(𝐶 2 + 𝐸𝑦 + 𝐹) ≥ 0. Rewrite (𝐵2 − 4𝐴𝐶)𝑦 2 + (2𝐵𝐷 − 4𝐴𝐸)𝑦 + 𝐷2 − 4𝐴𝐹 ≥ 0 If 𝐵2 − 4𝐴𝐶 = 0, then 1. If 2BD-4AE=0, the range is (−∞, ∞) if 𝐷2 − 4𝐴𝐹 ≥ 0 and ∅ otherwise. 2. The range is [− 𝐷2 −4𝐴𝐹 2𝐵𝐷−4𝐴𝐸 3. The range is (−∞, − , ∞) if 2𝐵𝐷 − 4𝐴𝐸 > 0 . 𝐷2 −4𝐴𝐹 2𝐵𝐷−4𝐴𝐸 ] if 2𝐵𝐷 − 4𝐴𝐸 < 0 . The discriminant is 𝑑1 = (2𝐵𝐷 − 4𝐴𝐸)2 − 4(𝐵2 − 4𝐴𝐶)(𝐷2 − 4𝐴𝐹) 1. If d1<0 , the range is (−∞, ∞) if 𝐵2 − 4𝐴𝐶 > 0 and ∅ otherwise. 2. If d1=0 , the range is (−∞, ∞) if 𝐵2 − 4𝐴𝐶 > 0 and {− 3. If d1>0 , there are two roots 𝑦1 = 4𝐴𝐸−2𝐵𝐷−√𝑑1 2(𝐵2 −4𝐴𝐶) 2𝐵𝐷−4𝐴𝐸 } otherwise. 2(𝐵2 −4𝐴𝐶) and 𝑦2 = 4𝐴𝐸−2𝐵𝐷+√𝑑1 2(𝐵2 −4𝐴𝐶) . The range is (−∞, 𝑦1 ] ∪ [𝑦2 , ∞) if 𝐵2 − 4𝐴𝐶 > 0 and [𝑦1 , 𝑦2 ] otherwise. Question #9 Find semi-major/semi-minor axis of ellipse from parametric equations with different phase 𝑥 = 𝑥̂ ⋅ cos(𝛺𝑡 − 𝜃) 𝑦 = 𝑦̂ ⋅ sin(𝛺𝑡 − 𝜑) Answer: Position vector 𝑟 = (𝑥, 𝑦) = (cos Ω 𝑡 − 𝜃, sin Ω 𝑡 − 𝜑) Define the variable 𝑡 ′ = Ω𝑡 − 𝜃, 𝑡ℎ𝑒𝑛 Ω𝑡 − 𝜑 = 𝑡 ′ + 𝜃 − 𝜑 = 𝑡 ′ + 𝜓 𝑤ℎ𝑒𝑟𝑒 𝜓 = 𝜃 − hence, 𝑟 = (𝑥, 𝑦) = (cos 𝑡 ′ , sin(𝑡 ′ + 𝜓)) Expanding the sine function, 𝑟 = (𝑥, 𝑦) = (cos 𝑡 ′ , sin 𝑡 ′ cos 𝜓 + cos 𝑡 ′ sin 𝜓) And this can be written as 𝑟 = (𝑥, 𝑦) = cos 𝑡 ′ (1, sin 𝜓) + sin 𝑡 ′ (0, cos 𝜓) x  y In matrix-vector form use a 2 × 1 vector for r, i.e. r =   , and let  cos t  u=   sin t   and let  1 V = sin 0  cos  Then r=Vu From which 𝑢 = 𝑉 −1 𝑟 Since 𝑢𝑇 𝑢 = cos 2 𝑡 ′ + sin2 𝑡 ′ = 1, then it follows that the algebraic equation of the ellipse is 𝑟 𝑇 𝑉 −𝑇 𝑉 −1 𝑟 = 1 Now all we have to do is find the eigenvalues of the matrix 𝑉 −𝑇 𝑉 −1 . We have that 1 cos V −1 = So,  cos  − sin  G = V −TV −1 = 1 cos2  0 1   1 − sin  − sin  1  We want to put G in the form: 𝐺 = 𝑅𝐷𝑅𝑇 Using standard diagonalization of a 2 × 2 matrix, and since 𝐺11 = 𝐺22 = 1, then the angle between the 𝜋 axes of the ellipse and the coordinate axes is 𝜙 = . 4 The diagonal matrix D entries are computed as follows D11 = G11 cos2  + G22 sin 2  + G12 sin 2 D22 = G11 sin 2  + G22 cos2  − G12 sin 2 These evaluate to, 𝐷11 = 𝐷22 = 1−sin 𝜓 cos2 𝜓 1+sin 𝜓 cos2 𝜓 = = 1 1+sin 𝜓 1 1−sin 𝜓 Therefore, the semi-major axis of the ellipse has length √1 + sin 𝜓 and the semi-minor axis has length √1 − sin 𝜓. Question #10 Find the parametric equation for the tangent line to the intersection curve between an ellipsoid and a paraboloid? We got various problems in this site asking for similar problems btw an ellipsoid and a plane. What if it's btw an ellipsoid and a paraboloid? I got the equation of both surfaces: 4𝑥 2 + 𝑦 2 + 𝑧 2 = 9 and 𝑧 = 𝑥 2 + 𝑦 2 respectively. Now given the tangent line touches their intersection (as an off-origin xz-ellipse but this might be useless) at (-1,1,2), how can we find the v- for the parametric equation of the tangent line 𝑟(𝑡) = (−1,1,2) + 𝑡𝑣 ? Is there any way we need to utilize the directional derivative of either surface? Or we just use f(x,y,z) and g(x,y,z) to describe both surfaces so that 𝑣 = ∇𝑓 × ∇𝑔 which I got after simplification as (5,8,6)? Thanks in advance. Answer: Step 1 You can parametrize the intersection curve of both surfaces and then find the direction vector of the tangent line. But the easier approach here is what you have written towards the end of the question. Let me try and address your question on how we know that tangent planes will intersect in the tangent line. Please note that tangent lines to level curves of both surfaces passing through point P(-1,1,2) will lie in the tangent planes to the surfaces at P. As the intersection curve is on both surfaces and passed through P, its tangent line at P must be on both tangent planes. A line can only lie in two different planes if both planes intersect and it must be the line that is intersection of both planes. Now coming to the first approach I mentioned. Given surfaces are, 𝑧 = 𝑥 2 + 𝑦2 4𝑥 2 + 𝑦 2 + 𝑧 2 = 9 Step 2 At intersection, x =  1 1 9 − z − z2 , y =  4 z + z 2 − 9, z = z 3 3 Looking at P, we parametrize the intersection curve for 𝑥 ≤ 0, 𝑦 ≥ 0 . r (t ) = (− 1 1 9 − t − t2 , 4t + t 2 − 9, t ) 3 3 Point P is represented by t=2 . 2t + 1 r (t ) = ( 2 27 − 3t − 3t 5 8 1)r (2) = ( , ,1) 6 6 2 , 2t + 4 2 12t + 3t 2 − 27 , So we can take the direction vector of the line as (5,8,6) and equation of the tangent line to the intersection curve at P is, 𝑙: (−1,1,2) + 𝑠~(5,8,6) Question #11 Finding equation of a path in the plane y=z What is the easiest way to see that the path 𝑟: 𝑅 → 𝑅𝟛 : 𝑡 ↦ (sin 𝑡, cos 𝑡, cos 𝑡) traces out an ellipse in the plane y=z ? I think first rotating 𝑅𝟛 by 𝜋 4 about the x-axis will help but I am not sure how to proceed. Answer: Explanation: The rotation by 𝜋 4 will yield the path (𝑥 ′ , 𝑦 ′ , 𝑧 ′ ) = (sin 𝑡 , √2 cos 𝑡 , 0) in the new coordinates. How do you eliminate t? Use sin2 𝑡 + cos 2 𝑡 = 1: 𝑥 ′2 + 𝑦 ′2 2 =1 Question #12 Finding the axis and orientation of an ellipse with matrices So I have this ellipse equation: 5𝑥 2 + 10𝑦 2 − 12𝑥𝑦 = 14 I'm asked to get the lengh of the semi-major and semi-minor axis, and it's orientation. Answer: Given 5𝑥 2 + 10𝑦 2 − 12𝑥𝑦 = 14 When this is translated into matrix-vector form, we define the position vector as  x p =   , then we write  y the equation as 𝑝𝑇 𝐴𝑝 + 𝑏 𝑇 𝑝 + 𝑐 = 0 Where A is a symmetric 2 × 2 matrix, b is a 2 × 1 vector, and c is a scalar. The first term is 𝑝𝑇 𝐴𝑝 = 𝑥 2 𝐴11 + 𝑦 2 𝐴22 + 2𝑥𝑦𝐴12 Comparing with the given equation, one finds that 5 A= −6 −6 10  The second term is 𝑏 𝑇 𝑝 = 𝑏1 𝑥 + 𝑏2 𝑦 Comparing with the given equation, we find that b is the zero vector. Finally, taking 14 to the left hand side, we deduce that c=-14 Therefore, the equation in matrix-vector form is 𝑝𝑇 𝐴𝑝 + 𝑐 = 0 The next thing you need to do is diagonlize matrix A, i.e. find a rotation matrix R and a diagonal matrix D such that 𝐴 = 𝑅𝐷𝑅𝑇 There is a standard way to do this diagonalization that you should memorize 1 2𝐴12 2 𝐴11 −𝐴22 1. Calculate 𝜃 = tan−1 cos  R=  sin   D11 3. Calculate the diagonal matrix D =   0 − sin   cos   2. Calculate the rotation matrix 0  where D22  D11 = A11 cos 2  + A22 sin 2  + A12 sin 2 D22 = A11 sin 2  + A22 cos 2  − A12 sin 2 1 −12 2 −5 Following the above steps, we find that 𝜃 = tan−1 Therefore 𝜃 (and (2𝜃) are in the first quadrant. Using the trigonometric identities 1 cos 2 𝜃 = (1 + cos 2 𝜃) 2 Since tan(2𝜃) = 12 5 1 18 3 2 13 √13 Hence cos 𝜃 = √ ( ) = 1 8 𝑎𝑛𝑑 sin 𝜃 = √ ( ) = 2 13 2 √13 Thus for the second step, we have the rotation matrix as R= 1 3  13  2 −2 3  For the third step, we have for diagonal matrix 9 4 6 13 13 13 𝐷11 = 5 ( ) + 10 ( ) − 6(2) ( ) = 1 4 9 6 13 13 13 𝐷22 = 5 ( ) + 10 ( ) + 6(2) ( ) = 14 With all these calculations, we can now write the equation of the ellipse as 𝑝𝑇 𝑅𝐷𝑅𝑇 + 𝑐 = 0 To put this in the standard form divide by (-c) = - (-14) = 14 , then 𝑝𝑇 𝑅𝐸𝑅𝑇 𝑝 = 1 where the matrix E = (D/14) is equal to 1 E = D /14 = 14  0  0  1 Now define the vector 𝑞 = 𝑅𝑇 𝑝 so that p=Rq , then it follows that 𝑞 𝑇 𝐸𝑞 = 1 Hence 𝑞12 𝐸11 + 𝑞22 𝐸22 = 𝑞12 14 + 𝑞22 1 =1 Thus the coordinate vector of q lies on an ellipse (in standard orientation) with semi-major axis = √14 , and semi-minor axis = √1 = 1 . Our ellipse which is in terms of the vector p is just a rotation of q-ellipse by the angle 𝜃 because p=Rq , and R is a rotation matrix by angle 𝜃 (counter clockwise). Question #13 Fitting a ballistic trajectory to noisy data where both spacial and temporal domains observations are noisy Fitting a curve to noisy data is somewhat trivial. However it generally assumes that data abscissa is fixed, and the error is computed on the ordinate. In my setup, I have 3D spacial observations of ballistic trajectories (that I model with a simple parabola), but the observations time are also noisy. Therefore, I have to estimate the initial position 𝑦0 , 𝑦0 , 𝑧0 and initial speed 𝑣𝑥0 , 𝑣𝑦0 , 𝑣𝑧0 , based on 4D (noisy) observations (𝑋𝑖 , 𝑌𝑖 , 𝑍𝑖 , 𝑇𝑖 ), 𝑖 ∈ [0, 𝑁] , such that they fit the following model:   x(t ) = x0 + vx t 0   y (t ) = y0 + v y0 t   z (t ) = z0 + vz t − g t 2 0  2 with t monotonically increasing with i. I'm not sure how to formulate such optimization problem because I have 6 parameters to estimate, but also 4N variables with only 3N equations… My intuition tells me there's only one single parabola that minimizes the error (MSE for example), but I can't formulate the problem. Answer: Hint. Define an error function like 2 2 𝑔 𝐸(𝑥0 , 𝑦0 , 𝑧0 , 𝑣𝑥0 , 𝑣𝑦0 , 𝑣𝑧0 ) = ∑𝑘(𝑥𝑘 − 𝑥0 − 𝑣𝑥0 𝑡𝑘 ) + (𝑦𝑘 − 𝑦0 − 𝑣𝑦0 𝑡𝑘 ) + (𝑧𝑘 − 𝑧0 − 𝑣𝑧0 𝑡𝑘 − 𝑡𝑘2 ) 2 2 so the minimum can easily be obtained by solving ∇𝐸 = 0 Question #14 Geometric meaning of ||𝑧 − 𝑧1 | − |𝑧 − 𝑧2 || = 𝑎, where z , z1 , z2  Answer: One way to define a hyperbola is a set of points, such that for any point P of the set, the absolute difference of the distances 𝑃𝐹1 , 𝑃𝐹2 to two fixed points 𝐹1 , 𝐹2 (the foci) is constant or, using the complex-plane notation from your question, a set of points, such that for any point z of the set, the absolute difference c of the distances |𝑧 − 𝑧1 |, |𝑧 − 𝑧2 | to two fixed points 𝑧1 , 𝑧2 (the foci) is constant So yes, the locus is a hyperbola. The mistake you're making when you try to “bring the equation to canonical form” 𝑥2 𝑎2 − 𝑦2 𝑏2 = 1 is that that form assumes that the two foci are on the x-axis and equidistant from the origin, whereas an arbitrary hyperbola can have its foci anywhere. If you need an explicit equation, it may be helpful to break each complex value into its real (x) and imaginary (y) components. ||𝑧 − 𝑧2 | − |𝑧 − 𝑧1 || = 𝑐 ||(𝑥 + 𝑖𝑦) − (𝑥2 + 𝑖𝑦2 )| − |(𝑥 + 𝑖𝑦) − (𝑥1 + 𝑖𝑦1 )|| = 𝑐 ||(𝑥 − 𝑥2 ) + 𝑖(𝑦 − 𝑦2 )| − |(𝑥 − 𝑥1 ) + 𝑖(𝑦 − 𝑦1 )|| = 𝑐 |√(𝑥 − 𝑥2 )2 + (𝑦 − 𝑦2 )2 − √(𝑥 − 𝑥1 )2 + (𝑦 − 𝑦1 )2 | = 𝑐 √(𝑥 − 𝑥2 )2 + (𝑦 − 𝑦2 )2 − √(𝑥 − 𝑥1 )2 + (𝑦 − 𝑦1 )2 = ±𝑐 Now, if you have an equation of the form √𝑢 − √𝑣 = 𝑤, then doing some algebra gives you (𝑤 2 − 𝑢 − 𝑣)2 = 4𝑢𝑣 , eliminating the inconvenient √ signs. (𝑐 2 − (𝑥 − 𝑥2 )2 − (𝑦 − 𝑦2 )2 − (𝑥 − 𝑥1 )2 − (𝑦 − 𝑦1 )2 )2 = 4((𝑥 − 𝑥2 )2 + (𝑦 − 𝑦2 )2 )((𝑥 − 𝑥1 )2 + (𝑦 − 𝑦1 )2 ) Question #15 Given: 𝑦 = 𝑥 1.65 Could it be described as parabolic in shape, or does the equation have to have 𝑥 2 as its highest degree term? Answer: Not everything u-shaped is a parabola. A parabola has a very specific shape with particular special properties. For example, for each parabola there is a point called the "focus" and a line called the "directrix" and each point on the parabola is the same distance from the focus as it is from the directrix. This is similar to the way a circle has a center and every point of the circle is an equal distance from the center. A shape might look more or less circular, but if it doesn't have a center that is the same distance from each of its points, it isn't a circle. Curves such as 𝑦 = 𝑥 1.65 , don't have a focus and directrix that behave the way a parabola's do. They might look parabolic, but they are not parabolas. Note also that an equation may have a parabolic graph even if it isn't obvious. The answer below of Vítězslav Štembera says that the equation must have one of two specific forms, but that isn't correct. Any equation of the form 𝐴𝑥 2 + 2𝐵𝑥𝑦 + 𝐶𝑦 2 + 2𝐷𝑥 + 2𝐸𝑦 + 𝐹 = 0 will have a parabolic graph, if it is non-degenerate and if 𝐴𝐶 − 𝐵2 = 0. For example, if one takes the parabola 𝑦 = 𝑥 2 and rotates it by 45∘ the curve is still a parabola, with equation 𝑥 2 − 2𝑥𝑦 + 𝑦 2 − 𝑥√2 − 𝑦√2 = 0. Question #16 How can you prove that the midpoint of the square will be Origin? Vertices A,B,C,D and the midpoints of the sides of the square ABCD lie on 𝑥 2 𝑦 2 = 1. No other points of the square lies on the curve. How can you prove that the midpoint of the square will be Origin ? My approach: If I can show if (x,y) are the coordinates of A then (y, -x) will be so of B , I will be done. I supposed A lies on first quadrant , B lies on second quadrant etc.. Can anyone tell please me how to show it ? Answer: You can try to argue geometrically using symmetries or you can try to find the coordinates of the vertices of the square. Let (x,y) be a vertex of the square in the first quadrant with smallest x. Then consider the point (y,-x). This point lies on the curve (in the second quadrant) and asking that the midpont ((x+y)/2,(y-x)/2) lies on the curve yields the equation 𝑥 4 + 𝑦 4 = 18. Substitution of 𝑦 2 = 1/𝑥 2 yield as solution (take the positive x with the smallest value) that 𝑥 = √√5 − 2 and 𝑦 = √√5 − 2−1 . Now you can check that the vertices (𝑥, 𝑦), (𝑦, −𝑥), (−𝑥, −𝑦) and (−𝑦, 𝑥) form a square satisfying the requirements, and its midpoint is the origin. Question #17 How to find the coefficient a of a 𝑦 = 𝑎𝑥 2 parabola? Answer: As a first qualitative step we know that if the parabola opens upward a>0 otherwise if it opens downward a<0 For a quantitative evaluation we need at least to know the coordinates of a point (𝑥0 , 𝑦0 ) on the parabola with 𝑥0 ≠ 0 and then 𝑦0 = 𝑎𝑥02 ⇒ 𝑎 = 𝑦0 𝑥02 Question #18 How to solve system of equations 𝐴1 𝑥 2 + 𝐵1 𝑥𝑦 + 𝐶1 𝑦 2 + 𝐷1 𝑥 + 𝐸1 𝑦 + 𝐹1 = 0 𝑎𝑛𝑑 𝐴2 𝑥 2 + 𝐵2 𝑥𝑦 + 𝐶2 𝑦 2 + 𝐷2 𝑥 + 𝐸2 𝑦 + 𝐹2 = 0 How to express x through y? Answer: You have the following system of quadratic equations: 𝐴1 𝑥 2 + 𝐵1 𝑥𝑦 + 𝐶1 𝑦 2 + 𝐷1 𝑥 + 𝐸1 𝑦 + 𝐹1 = 0 and 𝐴2 𝑥 2 + 𝐵2 𝑥𝑦 + 𝐶2 𝑦 2 + 𝐷2 𝑥 + 𝐸2 𝑦 + 𝐹2 = 0 First, transform each of the given quadratic equations into quadratic form notation as follows. Define  x r =  y  , then the first equation can be written as the quadratic form 𝑟 𝑇 𝑄1 𝑟 = 0 where  1    A1  B Q1 =  1 2 D  1  2 B1 2 C1 E1 2 D1  2  E1  2  F1   Similarly, the second quadratic equation can be expressed as the quadratic form 𝑟 𝑇 𝑄2 𝑟 = 0, with   A2  B Q2 =  2  2 D  2  2 D2  2   E2  2   F2   B2 2 C2 E2 2 Points of intersection of the two equations will satisfy 𝑟 𝑇 𝑄1 𝑟 = 0 and 𝑟 𝑇 𝑄2 𝑟 = 0 Therefore, the points of intersection will satisfy 𝑟 𝑇 (𝛼𝑄1 + 𝛽𝑄2 )𝑟 = 0 Let 𝑄 ∗ = 𝛼𝑄1 + 𝛽𝑄2 , we want to make the determinant of Q∗ zero for reasons that will become clear promptly. And to simplify the expression for the determinant, we'll consider first the case where 𝛼 = 0 , if det(𝑄2 ) = 0 , then we're done, otherwise, we can assume that 𝛼 ≠ 0 , and can therefore write 𝛽 𝑄 ∗ = 𝛼 (𝑄1 + 𝑄2 ) 𝛼 Taking 𝛼 = 1 is as good as any non-zero value for α, this simplifies Q∗ to 𝑄 ∗ = 𝑄1 + 𝛽𝑄2 Since Q1 and Q2 are 3 × 3 symmetric matrices, then det(Q∗) will be a cubic polynomial in β, which must have at least one real root. Once we substitute this root into Q∗ we have our rank deficient matrix Q∗. A Rank deficient square matrix has at least one of its eigenvalues equal to zero, and this leaves the other two eigenvalues as both positive , both negative, or one positive and one negative. Diagonalizing Q∗ we obtain 𝑄 ∗= 𝑅𝐷𝑅𝑇  D11  where D = 0   0 0 D22 0 0 0 0 Remember that our equation which we are trying to solve is 𝑟𝑇 𝑄∗𝑟 = 0 therefore, we want to solve 𝑟 𝑇 𝑅𝐷𝑅𝑇 𝑟 = 0 Define 𝑤 = 𝑅𝑇 𝑟, 𝑖. 𝑒. 𝑟 = 𝑅𝑤 , then we end up with 𝑤 𝑇 𝐷𝑤 = 0 Recalling what D is, this is 𝐷11 𝑤12 + 𝐷22 𝑤22 = 0 Now if both 𝐷11 and 𝐷22 have the same sign, then the solution is 𝑤1 = 𝑤2 = 0 while 𝑤3 can be any real number. In this case, since r=Rw must have its third coordinate equal to 1, then this case corresponds to a single point. We're not done, we have to verify that point we obtained satisfies 𝑟 𝑇 𝑄1 𝑟 = 0 If it does then this is our solution (a single point), otherwise there are no solutions. Next, we'll consider the case where the eigenvalues have unequal signs, then we can choose that 𝐷11 > 0 and 𝐷22 < 0 . In this case, the equation in vector w becomes 𝐷11 𝑤12 + 𝐷22 𝑤22 = 0 so that 𝑤2 = ±√− 𝐷11 𝐷22 𝑤1 while 𝑤3 can be any real number. Thus in this case 𝑤 = 𝑡(0,0,1) + 𝑠 (1, ±√− 𝐷11 𝐷22 , 0) That is w lies in two possible planes spanned by the above two vectors for each of the two possible signs. Now r=Rw Again, a plane in the r space for each choice of the sign in the above equation. Intersecting each plane (in x,y,z of vector r ) with the plane z=1 (because we want the third coordinate of r to be 1 ), we obtain exactly two lines of intersection. Each line is of the form r = r0 +  d0 ,   For each of these lines, we now enforce 𝑟 𝑇 𝑄1 𝑟 = 0 This will ensure that 𝑟 𝑇 𝑄2 𝑟 = 0 as well, because we know that r satisfies 𝑟 𝑇 (𝑄1 + 𝛽𝑄2 )𝑟 = 0 . This will give at most two solutions for the resulting quadratic equation in 𝜆. Thus we can have at most four solutions satisfying our system. Question #19 I have the general equation 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 and I want to determine what conic section it represents according to the value of its coefficients or relations between them, is there a theorem where that is explained? Answer: This equation can be written in matrix notation, in terms of a symmetric matrix to simplify some subsequent formulae, as  A B / 2 T T Q = ( x, y)   ( x, y) + ( D E )( x, y) + F = 0. B/2 C  the first matrix X is called the matrix of the quadratic form. Q is a hyperbola if and only if det A<0 , Q is a parabola if and only if det A=0 , and Q is an ellipse if and only if det A>0 . Question #20 If x and y are real numbers and 4𝑥 2 + 2𝑥𝑦 + 9𝑦 2 = 100 then what are all possible values of 𝑥 2 + 2𝑥𝑦 + 3𝑦 2 . Answer: Hint: The problem amounts to minimizing and maximizing 𝑥 2 + 2𝑥𝑦 + 3𝑦 2 subject to the constraint 4𝑥 2 + 2𝑥𝑦 + 9𝑦 2 = 100 . The possible values of 𝑥 2 + 2𝑥𝑦 + 3𝑦 2 will be the closed interval from the minimum to the maximum. Finding the necessary minimum and maximum can then be tackled using the method of Lagrange multipliers.