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Conic Sections Imagine you slice through a cone at different angles You could get a cross-section which is a: circle ellipse parabola rectangular hyperbola These shapes are all important functions in Mathematics and occur in fields as diverse as the motion of planets to the optimum design of a satellite dish. In FP1 you consider the algebra & geometry of 2 of these – the parabola and rectangular hyperbola The Parabola Parabola.agg Consider a point P that can move according to a rule: Q is the point horizontally in line horizontally with P on the line x = -a The point S has coordinates (a,0) P Q P can move such that QP=PS … The locus of points for P is a parabola The point S(a,0) is called the focus The line x = -a is called the directrix S(a,0) The restriction that P can move such that QP = PS is the focus-directrix property The Cartesian equation is y2 = 4ax x = -a WB14 Figure 1 shows a sketch of the parabola C with equation y 2 36 x (a) The point S is the focus of C. Find the coordinates of S. y 2 4ax where the focus is S(a,0) Figure 1 and the directrix has equation x = -a 25 16 9 Coordinates of S are (9,0) 24 (b) Write down the equation of the directrix of C. Equation of directrix x = -9 9 Figure 1 shows the point P which lies on C, where y > 0, and the point Q which lies on the directrix of C. The line segment QP is parallel to the x-axis. (c) Given that the distance PS is 25, write down the distance QP, Focus-directrix property: PS = PQ QP = 25 (d) find the coordinates of P, Sub x 16 in y 2 36 x y 24 Coordinates of P are (16,24) (e) find the area of the trapezium OSPQ. Area 9 25 24 408 2 WB15 Figure 1 shows a sketch of part of the parabola with equation y2 = 12x . 1 The point P on the parabola has x-coordinate 3 Figure 1 The point S is the focus of the parabola. (a) Write down the coordinates of S. y 2 4ax where the focus is S(a,0) 3 31 3 31 2 Coordinates of S are (3,0) 3 3 The points A and B lie on the directrix of the parabola. The point A is on the x-axis and the y-coordinate of B is positive. Given that ABPS is a trapezium, (b) calculate the perimeter of ABPS. Directrix has equation x = -a AO OS 3 Sub x 1 3 in y 2 12 x y 2 at P AB 2 Focus-directrix property PS PB 3 1 3 Perimeter = 14 32 Parametric functions Some simple-looking curves are hard to describe with a Cartesian equation. Parametric equations, where the values of x and y are determined by a 3rd variable t, can be used to produce some intricate curves with simple equations. Eg a curve has parametric equations x 2t , y t Complete the table and sketch the curve t -3 -2 -1 0 1 2 3 x -6 -4 -2 0 2 4 6 y 9 4 1 0 1 4 9 NB: you can still find the Cartesian equation of a function defined parametrically… x 2t t x 2 2 x 1 2 2 Sub in y t y x 2 4 2 Problem solving with parametric functions Eg a curve has parametric equations x t 1 , y 4 t 2 The curve meets the x-axis at A and B, find their coordinates 2 At A and B, y 0 t 4 Find values of t at A and B Substitute values of t back into expression for x A B t 2 x 3,1 Coordinates are (-3,0) and (1,0) Eg a curve has parametric equations x t , y 4t 2 The line x y 4 0 meets the curve at A. Find the coordinates of A t 2 4t 4 0 t 2 0 Substitute the expressions for x and y in terms of t to solve the equations simultaneously 2 t 2 A4,8 Solve Substitute value of t back into expressions for x and y The parametric form of a parabola is x at , y 2at Does this fit with its Cartesian equation? 2 4a t 2 Sub into y 4ax 2at 4a at 2 2 2 2 4a 2t 2 which is true! Exam questions sometimes involve the parabola’s parametric form… WB16 The parabola C has equation y2 = 20x. (a) Verify that the point P(5t2 ,10t) is a general point on C. 100t 2 Sub 5t ,10t in y 20 x 10t 20 5t 2 2 2 2 100t 2 The point A on C has parameter t = 4. The line l passes through A and also passes through the focus of C. (b) Find the gradient of l. A80,40 40 t 4 A80,40 S 5,0 75 y 4ax has focus S(a,0) 2 y 2 20 x a 5 S 5,0 40 Gradient of l 75 158 l WB17 The parabola C has equation y2 = 48x. The point P(12t2, 24t) is a general point on C. (a) Find the equation of the directrix of C. y 2 4ax where the focus is S(a,0) and the directrix has equation x = -a Equation of directrix x = -12 (b) Show that the equation of the tangent to C at P(12t2, 24t) is x − ty + 12t2 = 0. dy 21 y 48 x y 4 3 x 2 3 x 2 dx dy 2 2 123t 2 1t at P Sub x 12t dx 1 2 2 The equation of the straight line with gradient m that passes through ( x1 , y1 ) is y y1 m( x x1 ) 3 x Giving tangent The tangent to C at the point (3, 12) meets the directrix of C at the point X. (c) Find the coordinates of X y 24t 1t ( x 12t 2 ) ty 24t 2 x 12t 2 x ty 12t 2 0 Directrix x 12 3,12 Comparing (3,12) with (12t2, 24t) t at (3,12) x 21 y 3 0 When this intersects directrix x = -12 Sub t X 1 2 1 2 n equation of tangent 12 21 y 3 0 y 18 Coordinates of X are (-12,-18) The Rectangular Hyperbola The rectangular hyperbola also has a focus-directrix property, but it is beyond the scope of FP1. You only need to know that: The Cartesian equation is xy = c2 The parametric form of a parabola is x ct , y c t Problems involving rectangular hyperbola usually require to find the equation of the tangent or normal for functions given explicitly or in terms of c dy c2 2 2 xy c y c x c x 2 dx 2 x dy c 1 Sub x ct 2 2 2 dx ct t 2 2 1 WB19 The rectangular hyperbola H has equation xy = c2, where c is a positive constant. The point A on H has x-coordinate 3c. (a) Write down the y-coordinate of A. (c) The normal to H at A meets H xy c 2 with general point ct, ct t 3 y coordinate c 3 (b) Show that an equation of the normal to H at A is 3 y 27 x 80c again at the point B. Find, in terms of c, the coordinates of B. Solve xy c 2 and 3 y 27 x 80c simultaneously to find points of intersection 2 dy c 2 2 c x 2 xy c y c x dx x 2 dy c 1 Sub 3c, 3c at A 2 dx 3c 9 2 1 2 Gradient of normal 9 The equation of the straight line with gradient m that passes through ( x1 , y1 ) is y y1 m( x x1 ) Giving normal y 3c 9( x 3c) 3 y c 27 x 81c 3 y 27 x 80c xy c y 2 Sub in c2 x 3 y 27 x 80c 3cx 27x 80c 2 3c 2 27 x 2 80cx 27 x 2 80cx 3c 2 0 Given solution x = 3c x 3c 27 x c 0 x 27c at B y 27c using y cx 2 Coordinates of B are 27c , 27c WB20 The point P 6t , 6t , t ≠ 0, lies on the rectangular hyperbola H with equation xy = 36. (a) Show that an equation for the tangent to H at P is y xy c 2 y c 2 x 1 Sub 6t , 6 t 2 dy c 2 2 c x 2 dx x 6 1 dy t at P 2 t dx 6t The equation of the straight line with gradient m that passes through ( x1 , y1 ) is y y1 m( x x1 ) Giving tangent y 6t t12 ( x 6t ) y 6t t12 x 6t y t1 x 12t 2 1 t2 x 12t (b) The tangent to H at the point A and the tangent to H at the point B meet at the point (−9, 12). Find the coordinates of A and B. Sub 9, 12 in y t12 x 12t 12 t92 12t 12t 2 9 12t 4t 2 4t 3 0 4t 2 2t 6t 3 0 2t 2t 1 32t 1 0 2t 32t 1 0 t 32 , 21 Sub in 6t , 6 t 9,4 , 3,6 WB18 The rectangular hyperbola H has equation xy = c2, where c is a constant. The point P ct, c t is a general point on H. (a) Show that the tangent to H at P has equation t2y + x = 2ct. xy c y c x 2 Sub ct, 2 c t 2 dy c 2 2 c x 2 dx x 1 dy c2 1 at P dx t2 ct 2 The equation of the straight line with gradient m that passes through ( x1 , y1 ) is y y1 m( x x1 ) Giving tangent y ct t12 ( x ct ) t 2 y ct x ct t y x 2ct 2 The tangents to H at the points A and B meet at the point (15c, –c). (b) Find, in terms of c, the coordinates of A and B. Sub 15c, c in t 2 y x 2ct ct 2 15c 2ct ct 2 2ct 15c 0 t 3t 5 0 t 3, 5 3c, 3c , -5c,- 5c Sub values in ct, c t Formulae sheet facts Standard form Parametric form Parabola Rectangular hyperbola y 2 4ax xy c 2 at ct , ct 2 , 2at Foci a , 0 Not required Directrices x a Not required Obtaining the gradient as a function of t Parabola a dy 21 y 4ax y 2 a x a x dx x 1 2 2 2 Sub x at dy a 1 dx at t Rectangular hyperbola 2 1 xy c y c x 2 c2 dy 2 2 c x 2 x dx dy c2 1 Sub x ct 2 2 2 dx ct t