Questions and Answers #2 Conic Sections

Questions and Answers Sheet 2 Conic Sections Question #1 Given the force field F, find the work required to move an object on the given oriented curve. F=<< y , x >> on the parabola 𝑦 = 2𝑥 2 from (0, 0) to (2, 8) Answer: F=<< y,x>> parabola 𝑦 = 2𝑥 2 (0,0) to (2,8) parameterizing 𝑟(𝑡) =≪ 𝑡, 2𝑡 2 ≫ ,0 ≤ 𝑡 ≤ 2 therefore r'(t)=<< 1,4t>> the work is 2 𝑤 = ∫ ≪ 2𝑡 2 , 𝑡 ≫⋅≪ 1,4𝑡 ≫ 𝑑𝑡 0 simplify 2 2 𝑤 = ∫ (2𝑡 2 + 4𝑡 2 )𝑑𝑡 = ∫ 6𝑡 2 𝑑𝑡 0 𝑤 = [2𝑡 3 ]20 = 2(2)3 = 16 0 Question #2 Find the area of the largest rectangle that can be inscribed in the ellipse 𝑥 2 /𝑎2 + 𝑦 2 /𝑏 2 = 1𝑥 Answer: The given elipse is 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1, whic means it is a 2a wide and 2b tall. The area of the rectangle is 𝐴 = (2𝑥)(2𝑦) = 4𝑥𝑦 We have to solve the ellipse equation for y, then substitute for y in the area. We can use just the positive square root, that represents the upper half of the ellipse 𝑦2 𝑏2 𝑦 𝑏 =1− 𝑥2 𝑎2 = √1 − 𝑦 = 𝑏 (√ 𝑥2 𝑎2 1 𝑎2 (𝑎2 − 𝑥 2 )) 𝑏 = √𝑎2 − 𝑥 2 𝑎 Substitute for y in the rectunge area equation. Then find the derivative 𝑏 𝐴 = 4𝑥𝑦 = 4𝑥 ( √𝑎2 − 𝑥 2 ) = 𝑎 𝐴′ = = = = 4𝑏 𝑎 4𝑏 4𝑏 𝑎 𝑎 ( 𝑥√𝑎2 − 𝑥 2 1 2 (√𝑎2 − 𝑥 2 − ( 𝑎 ((1)√𝑎2 − 𝑥 2 + 𝑥 × (𝑎2 − 𝑥 2 )−2 (−2𝑥)) (product rule) 𝑎2 −𝑥 2 √𝑎2 −𝑥 2 4𝑏 𝑎2 −2𝑥 2 𝑎 1 4𝑏 √𝑎2 −𝑥 2 − 𝑥2 √𝑎2 −𝑥 2 𝑥2 √𝑎2 −𝑥 2 ) Find where A'=0 0 = 𝑎2 − 2𝑥 2 2𝑥 2 = 𝑎2 ) ) (common denominator) 𝑥√2 = 𝑎 𝑎 𝑥= √2 That's a local maximum of A, and since the minimum is 0 that it's likely this is the only maximum. Find the corresponding y value with the ellipse equation: 𝑏 𝑦 = √𝑎2 − 𝑥 2 𝑎 𝑏 𝑎 = √𝑎 2 − ( ) 𝑎 √2 𝑏 = √𝑎 2 − 𝑎2 𝑎 𝑏 = √ 2𝑎2 𝑎 2 𝑏 𝑎2 𝑎 𝑏 2 𝑎 𝑎 𝑏 √2 = √ = × = 2 − 2 𝑎2 2 √2 Now, put x and y into the rectangle area equation 𝐴 = 4𝑥𝑦 = 4 × 𝑎 √2 × 𝑏 √2 = 4𝑎𝑏 2 = 2𝑎𝑏 Question #3 Can anyone help with paramaterization of conics? Im struggling to wrap my head around an example. It considers the conic 𝑥 2 + 𝑦 2 − 𝑧 2 = 0 then proceeds: Take A=[1,0,1] and the line P(U) defined by x=0. Note that this conic and the point and line are defined over any field since the coefficients are 0 or 1. A point 𝑋 ∈ 𝑃(𝑈) is of the form X=[0,1, t] or [0, 0, 1] and the map 𝛼 is  ([0,1, t ]) = [ B((0,1, t ), (0,1, t ))(1, 0,1) − 2 B((1, 0,1), (0,1, t ))(0,1, t )] = 1 − t 2 , 2t ,1 + t 2  or  ([0, 0,1]) = [ −1, 0,1]. How do I evaluate B(v,v) or B(v,v)(a,b,c) like they have to go from the first line to the second? Answer: For the first question, it is exactly evaluation of a bilinear form, given by the matrix B, on the given vectors: so 𝐵(𝑣, 𝑤) = 𝑣 𝑡 𝐵𝑤. This gives you a scalar which is understood multipling the following vector. For the second one, when you have a projective line l and two dinstinct points P,Q on it then every other point can be written as 𝜆𝑃 + 𝜇𝑄 , for [𝜆, 𝜇] ∈ 𝑃𝟙 . So you can define 𝑡 = 𝜇/𝜆, if 𝜆 ≠ 0 , and consider l as the points in the form P+tQ union the remaining point corresponding to 𝜆 = 0 . Question #4 Why tangents to a quadratic curve never cut it again? I have been studying conic sections lately and couldnt figure out why we can use the condition D=0 for the quadratic equation having only one root in many of the questions involving finding tangents, which I have read in certain places is called the envelope method. That got me to thinking why the tangent to the conic passes through one and only one point on the conic. A tangent is defined only as a line which just touches the curve at one point. However, it lays no restrictions on whether it could intersect the curve again or not. So why exactly does a tangent to a circle/ ellipse/ parabola/ hyperbola never cut it again? Answer: The "sophisticated" answer is known as "Bezout's theorem". If you have a curve of degree m and a curve of degree n, then they intersect in at most mn places. (A "curve of degree m" is a curve with its equation given by a polynomial of degree n.) So a conic (degree 2) and a line (degree 1) intersect at most 2 \times 1=2 times. But you have to be careful about how you count the intersections. It turns out that a tangent counts as two intersections. Roughly speaking, this is because if you move the tangent line a little you can make it intersect the conic twice. The conic has equation 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 and the line has equation y=mx+b assuming it's not a vertical line. So you can substitute y=mx+b into the equation for the conic and get 𝐴𝑥 2 + 𝐵𝑥(𝑚𝑥 + 𝑏) + 𝐶(𝑚𝑥 + 𝑏)2 + 𝐷𝑥 + 𝐸(𝑚𝑥 + 𝑏) + 𝐹 = 0. Then you get scared that expanding this out will be a pain. But you don't have to! Just observe that x never occurs with power greater than 2. So it's a quadratic in x, and has it has at most two solutions. If the line is a vertical line, then it's x=k or some constant k, and substituting this into the conic gives a quadratic in y. Question #5 Why is this solution incorrect? Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ 1 1 of the parabola be drawn through it, then 2 + 2 is the same for all positions of the chord. 𝑃𝐾 𝑄𝐾 If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for 𝑦 2 = 4𝑎𝑥. let the point K be (c,0) Equation of line PQ using parametric coordinates: 𝑥 = 𝑐 + 𝑟 cos 𝜃 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1), 𝑦 = 𝑟 sin 𝜃 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2) From equation 1 and 2: (𝑥 − 𝑐)2 + 𝑦 2 = 𝑟 2 (Equation 3) Using Equation 3 and 𝑦 2 = 4𝑎𝑥, we get this quadratic in r: 𝑟 2 − 4𝑎𝑥 − (𝑥 − 𝑘)2 = 0 (Equation 4) Roots of this quadratic ( 𝑟1 and 𝑟2 ) would be the lengths of PK and QK From Equation 4, 𝑟1 + 𝑟2 = 0 and 𝑟1 𝑟2 = −(4𝑎𝑥 + (𝑥 − 𝑘)2 ) We know, 1 𝑃𝐾 2 + 1 𝑄𝐾 2 = 1 𝑟12 + 1 𝑟22 = (𝑟1 +𝑟2 )2 −2𝑟1 𝑟2 (𝑟1 𝑟2 )2 As value of 𝑟1 𝑟2 is not constant thus 1 𝑃𝐾 2 + = 1 𝑄𝐾 2 −2 𝑟1 𝑟2 does not turn out to be constant. Hence, this solution is incorrect. I've seen the correct solution but I wanted to know why this solution is incorrect? Answer: Roots of this quadratic (𝑟1 and r2 )would be the lengths of PK and QK. This part is incorrect. Roots of this equation correspond to the same value of x. P and Q can have different x. Update: 𝑥 = 𝑐 + 𝑟 cos 𝜃 𝑦 2 = 4𝑎𝑥 ⇒ 𝑟 2 sin2 𝜃 − 4𝑎𝑟 cos 𝜃 − 4𝑎𝑐 = 0 Absolute values of roots of this equation would be the PK and QK. 𝑟1 + 𝑟2 = 1 𝑃𝐾 2 + 1 𝑄𝐾 2 4𝑎 cos 𝜃 sin2 𝜃 (𝑟1 +𝑟2 )2 −2𝑟1 𝑟2 = (𝑟1 𝑟2 )2 = 16𝑎2 cos2 𝜃+8𝑎𝑐 sin2 𝜃 16𝑎2 𝑐 2 This value does not depend on 𝜃𝑎𝑡 𝑐 = 2𝑎 . This is the answer. Question #6 Why do we always take the parametre of parametric equation of ellipse as the angle formed with x-axis instead of semi-major axis? For ellipses having equation in the form of 𝑥2 𝑎2 + 𝑦2 𝑏2 =1 why are the parametric equations are always 𝑥 = 𝑎 cos(𝜃) 𝑦 = 𝑏 sin(𝜃) even when b>a ? Answer: The parametrization of the ellipse with an equation 𝑥2 𝑎2 + 𝑦2 𝑏2 =1 will be  x = a cos    y = b sin  for \theta ∈ [0,2𝜋]. Given this equation, we now define that: 1. An ellipse is vertically oriented if b>a , and 2. An ellipse is horizontally oriented if b 1. # Interestingly, for the circle the focus is at the center and l= radius a and e=0. The equation is simply r=a. Here, the equation is 3 1 1 = 1 + cos 𝜃. So, 𝑒 = < 1, and so, the conic is an ellipse. The semi major axis a is given by 𝑟 2 2 𝑙 = 𝑎(1 − 𝑒 2) 1 𝑎 1 4 4 4 = 𝑎 (1 − ) = 3 = 3. So, 𝑎 = . Question #8 How do you name the curve given by the conic r=6 ? Answer: It is a circle with center at (0, 0) and radius 6. Explanation: r=6 denotes locus of a point which moves so that it's distance from origin is always constant and at 6. It is quite apparent that this is the equation of a circle with center at origin and radius 6. Further, relation between polar coordinates (𝑟, 𝜃) and Cartesian coordinates (x, y) is given by 𝑥 = 𝑟 cos 𝜃 and x = 𝑟 sin 𝜃 𝑖. 𝑒. 𝑟 2 = 𝑥 2 + 𝑦 2 . Hence, r=6 on squaring translates into 𝑥 2 + 𝑦 2 = 36 and is equation of circle with center at (0, 0) and radius is 6. Question #9 How do you find the points of intersection of the curves with polar equations 𝑟 = 6 cos 𝜃and 𝑟 = 2 + 2 cos 𝜃 ? Answer: I would substitute the first equation into the second for r, giving: 6 cos(𝜃) = 2 + 2 cos(𝜃) And solving for \theta you get: 6 cos(𝜃) − 2 cos(𝜃) = 2 4 cos(𝜃) = 2 1 cos(𝜃) = which is valid: for 𝜃 = 𝜋 3 2 and𝜃 = 5 𝜋 3 Substituting back you get: 1 𝑟 = 6 cos(𝜃) = 6 ⋅ = 3 2 1 or 𝑟 = 2 + 2 cos(𝜃) = 2 + 2 ⋅ =3 again. 2 Giving 2 points of intersection: 𝜋 (3, ) 3 5 (3, 𝜋) 3 Question #10 Where is the vertical directrix of a conic if the denominator of its polar equation has the form 1 − 𝑒 cos 𝑞 ? Answer: You need the numerator. If the polar equation of your conic is 𝑟 = 𝑝 , both directrix are vertical and one of them contains the 1+𝑒 cos(𝜃) 𝑃 point 𝑃 ( , 0). 𝑒 If the denominator is 1 − 𝑒 cos(𝜃), write: 1 − 𝑒 cos(𝜃) = 1 + 𝑒 cos(𝜃 − 𝜋). 𝑝 Both directrix are vertical and one of them contains the point 𝑃 (− , 0). 𝑒 Question #11 How do you find the eccentricity, directrix, focus and classify the conic section 𝑟 = 0.8 1−0.8 sin 𝜃 ? Answer: It is a typical equation of an ellipse in polar form. However, it is easier to identify conic section, its eccentricity, directrix and focus in rectangular coordinates. Hence, let us convert the polar equation in rectangular form. The relation between polar form (𝑟, 𝜃) and rectangular form (x, y) is given by 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 𝑖. 𝑒. 𝑟 2 = 𝑥 2 + 𝑦 2 . 0.8 Hence 𝑟 = can be written as 1−0.8 sin 𝜃 4 4 𝑟 − 𝑟 sin 𝜃 = 5 5 or 5𝑟 − 4𝑟 sin 𝜃 = 4 or 5√𝑥 2 + 𝑦 2 − 4𝑦 = 4 or 25𝑥 2 + 25𝑦 2 = (4 + 4𝑦)2 = 16𝑦 2 + 32𝑦 + 16 or 25𝑥 2 + 9𝑦 2 − 32𝑦 − 16 = 0 or 25𝑥 2 + 9 (𝑦 2 − 2 × or 25𝑥 2 + 9 (𝑦 − or or 𝑥2 400 9×25 𝑥2 4 2 ( ) 3 + + 16 2 ) 9 400 9×9 16 2 (𝑦− (𝑦− ) 9 20 2 ( ) 9 16 9 16 2 9 ) = 16 2 𝑦+( ) )− 9 256 9 − 16 = 0 400 9 =1 =1 Hence, this is the equation of an ellipse of the form parallel to y-axis is 2 × 20 9 = 40 (𝑥−ℎ)2 𝑎2 + (𝑦−𝑘)2 𝑏2 16 = 1, whose center is (0, ), major axis 9 4 8 3 3 and minor axis parallel to x-axis is 2 × = 9 eccentricity is given by 𝑒 = √1 − 𝑎2 𝑏2 4 2 ( ) 9 = √1 − 3 2 = √1 − = 0.8 25 20 ( ) 9 Focii are (ℎ, 𝑘 ± be) i.e. (0, i.e. 𝑦 = 41 9 16 9 ± 16 9 32 𝑏 16 9 𝑒 9 )i.e. (0,0) and (0, ) and directrix are 𝑦 = 𝑘 ± i.e. 𝑦 = and 𝑦 = −1. Question #12 How do you find the eccentricity, directrix, focus and classify the conic section 𝑟 = Answer: 𝜋 Eccentricity is 2, Focus is at the pole (0, ) 2 Directrix is p=1 unit at right from the pole. Conic: Hyperbola Explanation: 2 𝑟= . The equation is in the form 1+2 cos 𝜃 2 1+2 cos 𝜃 ? ± 25 9 𝑟= 𝑒𝑝 1+𝑒 cos 𝜃 ; e=2 since e>1 the conic is hyperbola. 𝑟 = 𝑒𝑝 1+𝑒 cos 𝜃 ; 𝑒𝑝 = 2 ∴ 𝑝 = 1 Directrix is at p=1 unit at right from the pole. 𝜋 Eccentricity is 2, Focus is at the pole (0, ) 2 Question #13 How do you find the eccentricity, directrix, focus and classify the conic section 𝑟 = 8 4−1.6 sin 𝜃 ? Answer: Eccentricity is e=0.4 , directrix is y=-5 , focus is at pole (0, 0) and the conic is ellipse. Explanation: 8 𝑟= , this is similar to standard equation, 𝑟= 4−1.6 sin 𝜃 𝑒𝑝 1−𝑒 sin 𝜃 𝑒, 𝑝 are eccentricity of conic and distance of directrix from the focus at pole. (-) sign indicates that the directrix is below the focus and parallel to the polar axis 𝑟= 𝑟= 8 4−1.6 sin 𝜃 8 4−1.6 sin 𝜃 = 8 4 4−1.6 sin 𝜃 4 or ∴ 𝑒 = 0.4 and 𝑒𝑝 = 2or𝑝 = 2 0.4 or 𝑝 = 5; 𝑒 < 1 ∴ conic is ellipse and directrix is 5 units below the pole parallel to the polar axis. Rectangular conversion: 4𝑟 − 1.6𝑟 sin 𝜃 = 8 or 4√𝑥 2 + 𝑦 2 − 1.6𝑦 = 8. Eccentricity is e=0.4 , directrix is y=-5, focus is at pole (0, 0) and the conic is ellipse. Question #14 How do I find the directrix of the parabola 𝑦 = 2𝑥 2 ? Answer: The eqn. of the directrix of the parabola 𝑥 2 = 4𝑏𝑦 is given by, y+b=0 . 1 As, in our Problem 𝑏 = , the desired eqn. is, 1 8 𝑦 = = 0, or, 8𝑦 + 1 = 0 . 8 Question #15 How do you find the eccentricity, directrix, focus and classify the conic section 𝑟 = Answer: 𝑙 𝑟 = 1 + 𝑒 cos(𝜃 + 𝛼) represents (parabola ellipse hyperbola) according as (e=<>1) Here, the form is 5 𝜋 = 1 − sin 𝜃 = 1 + cos (𝜃 + ). 𝑟 2 The eccentricity e=1 . So, the conis is a parabola. 5 The semi latus rectum 2a=5 . So, the size of the parabola 𝑎 = . 2 The focus is at the pole r=0 . 𝜋 The axis of the parabola makes an angle 𝜃 = 𝛼 = . 𝜋 2 The vertex V is in the opposite direction 𝜃 = − , at a distance 2 10 2−2 sin 𝜃 ? 5 5 𝜋 2 2 2 𝑎 = . So, V is ( , − ) The directrix is perpendicular to the axis at a distance 2a=5 above from the vertex. So, its equation is 𝑟 sin 𝜃 = −5. by projection of the radius to (𝑟, 𝜃) upon the axis, remembering that focus is at r=0 .. Question #16 How do you find the eccentricity, directrix, focus and classify the conic section 𝑟 = 14.4 2−4.8 cos 𝜃 ? Answer: We have 𝑎 𝑟= √𝑥 2 𝑏−𝑐 cos 𝜃 + 𝑦2 = or 𝑎 or 𝑥 𝑏−𝑐 √𝑥2 +𝑦2 √𝑥 2 + 𝑦 2 = 1= 𝑎√𝑥 2 +𝑦 2 𝑏√𝑥 2 +𝑦 2 −𝑐𝑥 𝑎 𝑏√𝑥 2 +𝑦 2 −𝑐𝑥 or or 𝑏√𝑥 2 + 𝑦 2 − 𝑐𝑥 = 𝑎 or 𝑥 2 + 𝑦2 = (1 − 𝑐2 𝑏2 (𝑐𝑥+𝑎)2 𝑏2 or ) 𝑥 2 + 𝑦2 − 2𝑎𝑐 𝑏2 𝑥= 𝑎2 𝑏2 so we have supposing 𝑏 ≠ 0 c=b parabola c>b huperbola c1 , the conic is named a hyperbola.. Here e=2 and 𝑙 = 4 = 𝑎(𝑒 2 − 1) = 3𝑎, and so, 4 the semi major axis the hyperbola 𝑎 = . 3 Question #19 How do you name the curve given by the conic 𝑟 = 4 1+cos 𝜃 ? Answer: Convert to the General Cartesian Form: 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 Compute the determinant: Δ = 𝐵2 − 4𝐴𝐶 If Δ < 0 , then it is an ellipse or a circle. If 𝐵 = 0and A = 𝐶, then it is a circle. Otherwise, it is an ellipse. If Δ = 0 , then it is a parabola. If Δ > 0 , them it is a hyperbola. Given: 𝑟 = 4 1+cos(𝜃) 𝑟 + 𝑟 cos(𝜃) = 4 Substitute 𝑟 = √𝑥 2 + 𝑦 2 and 𝑟 cos(𝜃) = 𝑥: √𝑥 2 + 𝑦 2 + 𝑥 = 4 √𝑥 2 + 𝑦 2 = 4 − 𝑥 𝑥 2 + 𝑦 2 = 𝑥 2 − 8𝑥 + 16 𝑦 2 + 8𝑥 − 16 = 0 Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, 𝐴 = 𝐵 = 𝐸 = 0, 𝐶 = 1, 𝐷 = 8, and 𝐹 = −16 Δ = 02 − 4(0)(1) = 0 It is a parabola. Question #20 A square ABCD has all it's vertices on 𝑥 2 𝑦 2 = 1. The midpoints of it's sides also lie on the same curve . What is the area of this square? It can be found very easily if we can show that diagonal of this square meet at the origin. Answer: If (𝑥0 , 𝑦0 ) is the centre of the square and the vector (2u, 2v) corresponds to one of the edges (𝑤𝑙𝑜𝑔. 𝑢 ≥ 𝑣 ≥ 0), then all the points (𝑥0 + 𝛼𝑢 + 𝛽𝑣, 𝑦0 + 𝛼𝑣 − 𝛽𝑣) with 𝛼, 𝛽 ∈ {−1,0,1} , except 𝛼 = 𝛽 = 0 , are on the curve. That is, (1) (𝑥0 + 𝛼𝑢 + 𝛽𝑣)(𝑦0 + 𝛼𝑣 − 𝛽𝑢) = ±1. (1) The only quadratic polynomials 𝐴𝑥 2 + 𝐵𝑥 + 𝐶 that take values ±1 for all 𝑥 ∈ {−1,0,1} are given by − 𝐴 = 𝐵 = 0, 𝐶 = ±1 − 𝐴 = ±2, 𝐵 = 0, 𝐶 = ∓1 − 𝐴 = 1, 𝐵 = ±1, 𝐶 = −1 − 𝐴 = −1, 𝐵 = ±1, 𝐶 = 1 For 𝛼 = 1 , (1) must turn into one of these in terms of 𝛽. As at least one of u, v is non-zero, at least one of the factors in (1) is non-constant and hence the first of the four options, degree 0, is excluded. It follows that the polynomial is of degree 2, i.e., both u and v are non-zero. In fact, we find that 𝑢𝑣 ∈ {1, 2} , which leaves us only with the possibilities −2𝛽 2 + 1, −𝛽 2 + 𝛽 + 1, −𝛽 2 + 𝛽 + 1. At any rate, (𝑥0 + 𝑢)(𝑦0 + 𝑣) = 1. The same argument for 𝛼 = −1 leads to (𝑥0 − 𝑢)(𝑦0 − 𝑣) = 1. By subtracting, we find (𝑥0 , 𝑦0 ) ⊥ (𝑣, 𝑢). Swapping the roles of 𝛼 and 𝛽, we find (𝑥0 + 𝑣)(𝑦0 − 𝑢) = −1 and (𝑥0 − 𝑣)(𝑦0 + 𝑢) = −1. This time, by subtracting, we find (𝑥0 , 𝑦0 ) ⊥ (−𝑢, 𝑣). As (𝑣, 𝑢) ⊥ (−𝑢, 𝑣), one of the three vectors must be zero, and the only candidate is (𝑥0 , 𝑦0 ), as desired. With that, 𝑦𝑣 = 1 and 𝑢 2 − 𝑣 2 = (𝑢 + 𝑣)(𝑢 − 𝑣) = 1, hence 𝑢2 is a root of 𝑥 2 − 𝑥 − 1, etc.