Module 3: Stand Up Conics

Transcription

1 MATH55 Module 3: Stand Up Conics Main Math concepts: Conic Sections (i.e. Parabolas, Ellipses, Hyperbolas), nd degree equations Auxilliary ideas: Analytic vs. Co-ordinate-free Geometry, Parameters, Calculus. Parabolas Problem a. A parabola is defined as the set of points equidistant from a given line (the directrix of the parabola) and a given point (the focus) which is not on the directrix. Let d > 0. Derive the equation of the parabola with directrix y = - d and focus (0, d). Modify for directrix y = d and focus ( 0, - d). Solution. (x,y) d d - d - d (x,y) For the first case, by the distance formula, (x 0) + (y d) = y + d. Squaring gives x + y dy + d = y + dy + d. Simplifying gives x = dy, or y = d x. For the second case, (x 0) + (y + d) = d y. The same steps give us y = d x. Summarizing, these parabolas are y = ±ax, where the positive parameter a = d. The midpoint between the focus and the directrix is the vertex ( the origin in these examples) and the line of symmetry through the vertex and perpendicular to the directrix is the axis of the parabola. Vertical (North-South) parabolas such as these can be translated to move the vertex to (h, k) as usual: y k = a(x h), or as a function y = a(x h) + k = ax ahx + (ah + k). Horizontal (East-West) parabolas with vertex at ( h, k )can be obtained by interchanging x and y : x h = a(y k), or as a function x = a(y k) + h = ay aky + (ak + h). In all cases, the common length from focus to vertex, and vertex to directrix, is d = a.

2 Problem b. Find vertex, focus, axis, intercepts, and sketch the graph for y + 6y x 5 = 0. Notice that it is horizontal, since the y term is squared. Solution. [Completing the square]. We complete the square in y: (y + 3y +?) (?) x 5 = 0. We want? = ( 3) in order to make a perfect square. Then: 0 = (y + 3y + 9 ) (9 ) x 5 = (y + 3 ) x 9 which we can now solve for x: x = (y + 3 ) 9, or x + 9 = (y + 3 ). Thus, the vertex is ( 9, 3 ), and the axis is the horizontal line y = 3. Now, a =, so it opens to the right, and the focal distance is d = a =, so the focus is ( 8, 3 ). Setting y = 0 gives x-intercept 5, and setting x = 0 gives y + 6y 5 = 0, and then the quadratic formula gives y- intercepts y = 6 ± 36 ()( 5) 6 ± 76 = = 3 ± , Solution. [Use Calculus to find the vertex]. We can solve for x, x = y + 3y 5. We immediately have a = and so d = ¼ and parabola opens right. The vertex is the point where dx/dy = 0, we can simply compute y + 3 = 0, y = 3 and then plug in to get the x co-ordinate x = ( 3 ) + 3( 3 ) 5 = 9, then the focus is ( 9 +, 3 ) = ( 8, 3 ). The intercepts can be found as is the first solution. y + 6y x 5 = axis y = - 3/ - 3

3 Problem c. A car is driving on toward the vertex on a parabolic road. There is a wall running along the axis of the parabola, and a laser pointer is mounted on the wall side the car, perpendicular to the car s direction. At what point on the wall is the laser pointing as the car reaches the vertex? Solution. Let the parabola open upwards with vertex (0,0). Then its equation is y = ax for some positive a. Since the laser is perpendicular to the car s path, the point on the wall where the laser shines is the y-intercept of the normal line to the parabola. So,if we let y N denote the y-intercept of the normal line at point P on the parabola, then we are seeking lim y. N P (0,0) At the point (x 0, y 0 ), on the parabola y = ax, the slope of the tangent line is y (x 0 ) = ax 0 and so the slope of the normal line is. Thus the equation of the normal line is ax 0 y y 0 = (x x 0 ). Since y 0 = ax 0, the normal line is y = x + ax 0 ax 0 a + ax 0. Then the y-intercept of the normal line is y N = 0 + ax 0 a + ax 0 = a + ax 0. Finally, as the car approaches the origin, we have lim y N = lim x 0 0 x 0 0 a + ax 0 = a So, if the car is driving on the parabola y = ax, the limiting point of the laser image on the wall is at a distance a from the vertex. Since the focus is at from the vertex, we could also state the result a (in a co-ordinate-free way) by saying the limiting point is twice as far from the vertex as is the focus. Picture shows case a = with the limiting intercept y = x P = (x 0, y 0 )

4 II. Ellipses and Hyperbolas Problem a. Consider the equation x 5 + y =, whose graph is an ellipse. 9 i) Find intercepts, x and y extents, symmetries, critical points and sketch the graph. ii) Let (x, y ) be a point on the graph.. Find the sum: the distance from (x, y) to (-,0) plus distance from (x, y) to (,0). Solution (i). The graph is symmetric about x-axis, y-axis, and origin since replacing x and/or y by its negation gives an equivalent equation. The intercepts are (±5,0) and (0,±3), and since both terms on the left side are non-negative and sum to, then neither term can be more than. So x 5 x 5 5 x 5 and similarly 3 y 3. Differentiating gives x 5 + y 9 y = 0, so y = x(9) 5(y) = 9x. This gives vertical and horizontal tangents at 5y (±5,0) and (0,±3) respectively, and also shows that the curve is decreasing in the first quadrant where x and y are both positive. By the symmetries, then, the graph must look like: x 5 + y 9 = Another way to view this ellipse is to consider the stretching transformation T(x,y)= (5x, 3y) which moves a point 5 times further from the y-axis and 3 times further from the x-axis. We apply T to the unit circle x + y = by substituting (x, y) T (x, y) = (x / 5, y / 3) to obtain the equation x 5 + y =. So this ellipse amounts to the unit circle 9 stretched by factors of 5 horizontally and 3 vertically. ii) The two distances whose sum we want to find are shown. By the distance formula, and using the fact that (x, y) is on the ellipse, we have the sum of the distances is: (x + ) + y + (x ) + y = (x + 8x + 6) + (9 (9 / 5)x ) + (x 8x + 6) + (9 (9 / 5)x ) (6 / 5)x + 8x (6 / 5)x 8x + 5 = (( / 5)x + 5) + (( / 5)x 5). Now we must be careful! Since 5 x 5, then - ( / 5)x, so ( / 5)x 5 is negative and (/5)x+5 is positive. So the desired sum is ( / 5)x + 5 (( / 5)x 5) = 0. So the sum is equal to 0 for all points on this ellipse. Thus an alternative definition of this ellipse is: The set of points (x, y ) whose sum of distances to (-,0) and (, 0) is 0. Geometrically, it s the set of points whose sum of distances, to two fixed points eight units apart, is 0.

5 The points ( -, 0) and (,0) are the foci of this ellipse. The center is (0,0) and the distances a=5 and b=3 are the semi-major axis and the semi-minor axis respectively. The common sum of distances is twice the semi-major axis (in our case, 0 = x 5). Notice that the distance from the center to a focus in our example is c =, and that a = b + c in our example. This is always the case, although we won t take the time for a complete derivation. We can interchange x and y to get a congruent ellipse centered at (0,0) whose major axis runs vertically, with foci are at (0,±) and intercepts (0,±5) and (±3,0). And of course, we can translate the ellipse to have ( h, k ) as its center. Summarizing: An ellipse is the set of points in a plane whose sum of distances to two given points, called the foci, is (x h) (y k) a given constant. The equation of an upright ellipse can be written as + = where: a b (h, k ) is the center (i.e. the midpoint between the foci), and we take a and b positive. The x-extent is (h a,h + a) and the y-extent is ( k b, k + b). The larger of a and b is the semi-major axis M, the smaller is the semi-minor axis m. The common sum of distances is M [either a or b]. (if a = b, then the ellipse is a circle of radius a = b ). (h ± c, k) where c = a b if a b The foci are at. The distance between foci is c. (h, k ± c) where c = b a if b a The eccentricity of the ellipse is e = c / M = c / (semi-major axis); then 0 e <. (eccentricity=0 is a circle, eccentricity near means very elongated) Problem b. Find the equation, and sketch, the set of points whose sum of distances to the origin and (0,6) is. Solution: We have a vertical (i.e. N-S) ellipse with foci (0,0) and (0,6), so the center is (0,3) and c = 3. For the major axis, b= so b = 6. Then a = b c = 36 9 = 7, so a = 3 3. So the equation is: x (y 3) + = ; the x extent is [ 3 3, 3 3], and the y extent is [-3,9] The intercepts are (±9 /,0), (0,-3), and (0,6). Eccentricity is 3 / 6 = x (y 3) + =

6 Problem c. Sketch the graph x 6 + y 9 asymptotes. = indicating interesting features such as symmetries and Solution. We have symmetry about the x-axis, y-axis and the origin by replacing variables with their negations.. Since x 6 = y 9, then y 9 0, i.e. y 9, so either y 3 or y 3. On the other hand, x can be any real number since we have y = ±3 + x /6 = ±3 x + 6 and any x will work. Also, we see that for large x, y will be like ± 3 x, so we suspect those lines may be asymptotes. Looking at the derivative, we have x / 6 + y y / 9 = 0, so y' = 9x. So there are horizontal tangents at x=0 and 6y the curve is increasing in the first quadrant. The graph, shows the curve as well as the lines y = ± 3 x.: 6 x 6 + y 9 = Lastly, we note that if we define the points (0,5) and (0,-5) as the foci of this hyperbola, then all points on the hyperbola have the difference of their distances from the foci equal to 6 (HW problem).

7 So a hyperbola can be defined geometrically as the set of points for which the difference of distances to two fixed points, called the foci of the hyperbola, is constant. As usual, we can interchange x and y to obtain a congruent East-West hyperbola, and we can translate the center to ( h, k ) as well. Combining everything, we have (x h) (y k) The equation of an upright hyperbola can be written as ± = (**) : a b (h, k ) is the center (i.e. the midpoint between the foci), and we take a and b positive. The distance between the foci is c, where c = a + b. The asymptotes go through the center and have slopes ± b a. The following statements refer to East West hyperbolas (i.e. x term is positive in (**)) The x-extent is (,h a] [h + a, ), the y extent is R. The length a from the center to the extreme points is the semi-major axis. The common difference of distances is a ( i.e. twice the semi-major axis). The eccentricity is c / a, which is always bigger than for a hyperbola. For North-South hyperbolas, interchange a and b, x and y, h and k in these four statements. III. Second Order polynomials Problem a. Consider the graph of the equation Ax + y x = 0 where A is a parameter. Determine those values of A for which it is an ellipse, parabola, hyperbola, and for those A, find its orientation. Solution. First notice that for any value of A, the curve goes through the origin (0, 0 ) and has a vertical tangent at (0,0) [vertical tangent verification is left to the reader] We try to put the equation in our standard form by completing the square. A(x A x + A ) + y = A A(x A ) + y =. Now multiply by A to get a on the right: A A (x A ) + Ay =. Finally write the coefficients in the denominators: (x / A) ( + y A ) ( A ) =. We let r = A, then we have (x r) r + y r =. First, if A > 0, then r > 0 and we have an ellipse centered at (r, 0) with a = r, b = r. For A > (i.e. r < ), a = r < r = b so the ellipses are N-S oriented. As A decreases (r increases) towards, both vertical and horizontal semi-axes grow, but the horizontal one grows faster ( r vs. r ) so the shape is more circular. At A = (i.e. r=), we have the circle of radius centered at (,0), and for 0 < A <, then r > and so a > b and hence the ellipses are oriented E-W.

8 As A decreases toward 0, r grows large and the ellipses become larger and more eccentric. The graph shows the ellipses for A =,, 0.5, 0.5, 0.5 ( i.e. r = /,,,, 0/3 ) y What happens when A reaches 0? The equation becomes y = x, a parabola with vertex at the origin, focus at (/,0), which opens to the right. Lastly, we look at the case where A < 0. Let S = - / A, so S > 0 and the equation is 0 = x S + y x = S (x + Sx + S ) + S + y Continuing, we get S (x + (x + S) + y S) = S, and hence y = which is standard form. S S These hyperbolas are East-West (since x-term is positive), centered at (-S, 0), semi-major axis S = /A, so the critical points are (0,0) and (-S, 0). The asymptotes have slope ±S / S = A. So, as A grows large negative, the hyperbolas become wider, and (as S approaches 0), the left piece moves closer to the origin. Ax + y x = 0 for A =,, 0.5, 0, -0.5, - hyperbolas (A < 0) 3 x parabola (A = 0) ellipses (A > 0) A picture of the parameter line and the behaviors for various values of A: A EW hyperbolas parabola EW ellipses circle NS ellipses