Math 114 - Conic Sections Peter A. Perry University of Kentucky April 13, 2017
Bill of Fare Why Conic Sections? Parabolas Ellipses Hyperbolas Shifted Conics
Goals of This Lecture By the end of this lecture, you should: Understand the geometric definitions of an ellipses, parabolas, and hyperbolas Know how to identify a given equation as the equation of an ellipse, a parabola, or a hyperbola Know how to find the equation of an ellipse parabola, or hyperbola from geometric information about the curve This lecture uses no Calculus. The only tools we will use are algebra, the Cartesian (x-y) coordinate system, and the distance formula.
A Glimpse into Your Future 10 0 5 0 2 1 x 0 z = x 2 + y 2 0 0 2 2 Traces of z = x 2 + y 2 x 0 1 2 2 y 2 y Next year, you will study functions of two variables, whose graph is a surface in three-dimensional space. Many of these graphs have cross sections (or traces) which are conic sections. For example, this surface, when cut by horizontal planes, has circles as cross sections. When cut by vertical planes, its cross sections are parabolas. The orbits of planets, comets, and stars are ellipses (closed orbit), hyperbolas (unbound orbit), or parabolas (unbound orbit) For these reason (among others), we ll study conic sections.
Why Section? Think of section as meaning slice Image from Wikipedia commons https://commons.wikimedia.org/wiki/file:conic_sections_2.png
Warm-Up: Circles r (h, k) (x, y) A circle is the set of all points in the xy plane at a given distance r from the center (h, k). The equation of a circle with center (h, k) and radius r is (x h) 2 + (y k) 2 = r 2 First comes the definition in words, then comes the equation of the conic
Ellipses At left is the graph of an ellipse with equation (0, b) x 2 a 2 + y 2 b 2 = 1, a b > 0 ( a, 0) ( c, 0) b a c (c, 0) (a, 0) But what makes an ellipse an ellipse? (0, b) a 2 = b 2 + c 2 Sum of distances is 2a An ellipse is the set of points in a plane the sum of whose distances from two fixed points, called the foci of the ellipse, is constant. Watch this video How do you get from that statement to an equation?
Ellipses Pick a point on the ellipse (0, b) P(x, y) ( a, 0) ( c, 0) (c, 0) (a, 0) (0, b)
Ellipses Pick a point on the ellipse Draw a lines from the point to the two foci (0, b) P(x, y) ( a, 0) ( c, 0) (c, 0) (a, 0) (0, b)
Ellipses Pick a point on the ellipse Draw a lines from the point to the two foci (0, b) P(x, y) Add up the distances to get 2a (x c) 2 + y 2 + (x + c) 2 + y 2 = 2a ( a, 0) ( c, 0) (c, 0) (a, 0) (0, b)
Ellipses Pick a point on the ellipse Draw a lines from the point to the two foci (0, b) P(x, y) Add up the distances to get 2a (x c) 2 + y 2 + (x + c) 2 + y 2 = 2a ( a, 0) ( c, 0) (c, 0) (a, 0) Square and simplify alot (a 2 c 2 )x 2 + a 2 y 2 = a 2 (a 2 c 2 ) (0, b)
Ellipses Pick a point on the ellipse Draw a lines from the point to the two foci (0, b) P(x, y) Add up the distances to get 2a (x c) 2 + y 2 + (x + c) 2 + y 2 = 2a ( a, 0) ( c, 0) (c, 0) (a, 0) Square and simplify alot (a 2 c 2 )x 2 + a 2 y 2 = a 2 (a 2 c 2 ) (0, b) Notice that a 2 c 2 = b 2 and get b 2 x 2 + a 2 y 2 = a 2 b 2 and then divide by a 2 b 2
Ellipses The ellipse shown has (0, b) semi-major axis a ( a, 0) a P(x, y) (a, 0) semi-minor axis b foci at (±c, 0) where ( c, 0) (c, 0) c 2 = a 2 b 2 (0, b) The sum of distances to the two foci is 2a The equation of the ellipse is x 2 a 2 + y 2 b 2 = 1
Ellipses If the semi-major axis lies on the y-axis, the picture changes as shown (0, a) (0, c) P(x, y) a The ellipse shown has semi-major axis a semi-minor axis b foci at (0, ±c) where ( b, 0) (b, 0) c 2 = a 2 b 2 (0, c) (0, a) The sum of distances to the two foci is 2a The equation of the ellipse is x 2 b 2 + y 2 a 2 = 1
(3, 0) (0, 2) Ellipses Recall for ellipse for semi-major axis a and semi-minor axis b, the foci at distance c from the origin where a 2 = b 2 + c 2 Example Find the equation of an ellipse with foci (0, ±2) and vertices (0, ±3) (0, 2) ( 3, 0)
( 5, 0) (3, 0) (0, 2) (0, 2) ( 5, 0) Ellipses Recall for ellipse for semi-major axis a and semi-minor axis b, the foci at distance c from the origin where a 2 = b 2 + c 2 Example Find the equation of an ellipse with foci (0, ±2) and vertices (0, ±3) First, complete the triangle using a = 3, c = 2 so 3 2 = b 2 2 2 or b = 5 ( 3, 0)
( 5, 0) (3, 0) (0, 2) (0, 2) ( 5, 0) Ellipses Recall for ellipse for semi-major axis a and semi-minor axis b, the foci at distance c from the origin where a 2 = b 2 + c 2 Example Find the equation of an ellipse with foci (0, ±2) and vertices (0, ±3) First, complete the triangle using a = 3, c = 2 so 3 2 = b 2 2 2 or b = 5 ( 3, 0) Knowing that a = 3 and b = 5, we get x 2 5 + y 2 9 = 1
What Do Ellipses Have to Do with Lithotripsy? Lithotripsy ( stone crushing ) is a medical procedure that crushes Kidney stones by tightly focussed ultrasound waves. Lithotripsy Machine Image via Wikipedia Commons The trick is to crush the stones noninvasively, without damaging the patient! The ultrasound generator lies at one focus of an ellipse, and the patient s kidneystones lie at the other focus.
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) F (0, p) (focus) y = p (directrix)
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) This is why telescope mirrors have a parabolic surface! (focus) y = p (directrix)
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) This is why telescope mirrors have a parabolic surface! F (0, p) (focus) y Translate this into an equation: Q p y = p (directrix)
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) This is why telescope mirrors have a parabolic surface! F (0, p) (focus) y Translate this into an equation: Q y = p (directrix) p x 2 + (y p) 2 = (y + }{{}}{{ p)2 } PF 2 QP 2
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) This is why telescope mirrors have a parabolic surface! F (0, p) (focus) y Translate this into an equation: Q y = p (directrix) p x 2 + (y p) 2 = (y + }{{}}{{ p)2 } PF 2 QP 2 x 2 + y 2 2py + p 2 = y 2 + 2py + p 2
Parabolas A parabola is a set of points equidistant from a fixed point, the focus, and a fixed line, the directrix P(x, y) This is why telescope mirrors have a parabolic surface! F (0, p) (focus) y Translate this into an equation: Q y = p (directrix) p x 2 + (y p) 2 = (y + }{{}}{{ p)2 } PF 2 QP 2 x 2 + y 2 2py + p 2 = y 2 + 2py + p 2 x 2 = 4py
Parabolas (0, p) y = p Parabola with focus (0, p) and directrix y = p x 2 = 4py x = p (p, 0) Parabola with focus (p, 0) and directrix x = p y 2 = 4px
Parabolas Recall x 2 = 4px or y 2 = 4px Focus is (0, p) or (p, 0), directrix is y = p or x = p Example Find the focus and directrix of a parabola whose equation is y 2 + 10x = 0
Parabolas Recall x 2 = 4px or y 2 = 4px Focus is (0, p) or (p, 0), directrix is y = p or x = p Example Find the focus and directrix of a parabola whose equation is y 2 + 10x = 0 y 2 = 10x
Parabolas Recall x 2 = 4px or y 2 = 4px Focus is (0, p) or (p, 0), directrix is y = p or x = p ( 5 2, 0) x = 5 2 Example Find the focus and directrix of a parabola whose equation is y 2 + 10x = 0 y 2 = 10x p = 5/2 Focus is at ( 5/2, 0), directrix is x = 5/2
Hyperbolas y P(x, y) A hyperbola is the set of all points the difference of whose distances from two fixed points (the foci) is constant. F 1 ( c, 0) F 2 (c, 0) x PF 1 PF 2 = ±2a
Hyperbolas y = b a x y y = b a x P(x, y) A hyperbola is the set of all points the difference of whose distances from two fixed points (the foci) is constant. This one has equation F 1 ( c, 0) ( a, 0) (a, 0) F 2 (c, 0) x where x 2 a 2 y 2 b 2 = 1 a 2 + b 2 = c 2 and has asymptotes PF 1 PF 2 = ±2a y = ± b a x
Hyperbolas y = b a x y y = b a x Hyperbolas come in two flavors, shown at left: x x 2 a 2 y 2 b 2 = 1 (foci (±c, 0), vertices (±a, 0), asymptotes y = ±(b/a)x) y = a b x y y = a b x and x y 2 a 2 x2 b 2 = 1 (foci (0, ±c), vertices (0, ±a), asymptotes y = ±(a/b)x)
Hyperbolas Example Sketch the graph of the equation y 2 25 x2 9 = 1 and find the foci of the hyperbola
Hyperbolas Example Sketch the graph of the equation (0, 5) y 2 25 x2 9 = 1 and find the foci of the hyperbola If x = 0 then y = ±5, so vertices at (0, ±5), so the hyperbolas open up and down (0, 5)
Hyperbolas Example Sketch the graph of the equation (0, 6) (0, 5) (0, 5) (0, 6) y 2 25 x2 9 = 1 and find the foci of the hyperbola If x = 0 then y = ±5, so vertices at (0, ±5), so the hyperbolas open up and down From the equation, a = 5 and b = 3 so c 2 = 25 + 9 = 36 or c = 6 This means the foci are at (0, ±6)
Which is Which? A. x 2 + y 2 = 4 B. y 2 x 2 = 1 C. x2 16 + y 2 9 = 1 D. x2 = 2y
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( 9 x 2 2x ) + 4y 2 = 36
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36 9 (x 1) 2 + 4y 2 = 36
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36 9 (x 1) 2 + 4y 2 = 36 (x 1) 2 4 + y 2 9 = 1
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36 9 (x 1) 2 + 4y 2 = 36 (x 1) 2 4 + y 2 9 = 1
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36 9 (x 1) 2 + 4y 2 = 36 (x 1) 2 What is the curve, and where is its center? 4 + y 2 9 = 1
Shifts So far, all of the curves have had center (0, 0). By completing the square we can analyze equations of conic sections whose center isn t at (0, 0). Example Identify and graph the equation 9x 2 18x + 4y 2 = 36 ( ) 9 x 2 2x + 4y 2 = 36 ( ) 9 x 2 2x + 1 + 4y 2 = 36 9 (x 1) 2 + 4y 2 = 36 (x 1) 2 What is the curve, and where is its center? 4 + y 2 9 = 1 Ellipse, center (1, 0), semi-major axis 3, semi-minor axis 2
Shifts Which one of the curves at left is the graph of (x 1) 2 4 + y 2 9 = 1?
Shifts Which one of the curves at left is the graph of (x 1) 2 + 9 (y 2)2 4 = 1?