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FP1 Coordinate Systems CONIC SECTIONS

FP1 Coordinate Systems CONIC SECTIONS. If we cut a cone at different angles, then we will obtain different types of conic section. There are four different types we can obtain. Perpendicular to the axis of the cone. This gives a CIRCLE.

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FP1 Coordinate Systems CONIC SECTIONS

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  1. FP1 Coordinate Systems CONICSECTIONS

  2. If we cut a cone at different angles, then we will obtain different types of conic section. There are four different types we can obtain. Perpendicular to the axis of the cone. This gives a CIRCLE.

  3. Cut at an angle to the axis so we still get a closed curve. This is an ELLIPSE

  4. Making a cut parallel to the slant height of the cone produces a PARABOLA.

  5. Make a cut even steeper and imagine we have a double cone, That is two cones vertex to vertex, then we obtain the two branches of a HYPERBOLA.

  6. Suppose we are given a fixed straight line, called a DIRECTRIX, and a fixed point, called the FOCUS, F. If we have another point P. We can consider the perpendicular distant to the line, MP, and the distance of P from the focus, FP What happens if one of these distances is a fixed multiple of the other? P M F focus directrix

  7. Suppose that, for some constant, e, the equation • PF = ePM • is ALWAYS true. • All the points P satisfying this equation lie on a curve called the LOCUS. • You will get different curves depending upon the value of the constant e: • If 0 < e < 1, then the curve is an ELLIPSE; • If e = 1 then the curve is a PARABOLA; • If e > 1 then the curve is a HYPERBOLA. • The constant e is called the ECCENTRICITY • of the conic. It is a measure of how far the • curve deviates from a circle. P M F focus directrix

  8. For a parabola, e = 1, so PF = PM THE PARABOLA This is called the FOCUS-DIRECTRIX PROPERTY for a parabola y a x P(x, y) PM = x + a M Using Pythagoras, PF = y So, x O F(a, 0) A(-a, 0) Which simplifies to give the standard Cartesian equation of a parabola:

  9. EXAMPLE 1 A parabola C has equation y2 = 6x. Find the coordinates of the focus of C and the equation of its directrix. Find the points where the line with equation y = 2x – 6 intersects C. So a = 3/2 4a = 6 (a) Compare y2 = 6x with y2 = 4ax Focus at (3/2, 0) Directrix has equation x = -3/2 (b) Substitute y = 2x – 6 into y2 = 6x (2x – 6)2 = 6x x = 3/2 and x = 6 4x2 – 24x + 36 = 6x y = -3 and y = 6 4x2 – 30x + 36 = 0 y = 2x – 6 intersects y2 = 6x at (3/2, -3) and (6, 6) 2x2 – 15x + 18 = 0 (2x – 3)(x – 6) = 0

  10. EXAMPLE 2 The diagram shows a parabola W, with focus F and directrix L. point P on W is such that the line FP makes an angle of 80° with the positive x-axis. Point Q on L is such that the line PQ is parallel to the x-axis. Find the angle PFQ. y W Q P 80° θ P(x, y) Q θ 80° x x O F 80° θ = ½ (180° – 80°) F θ = 50° L

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