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Coordinate Systems and Parametric Equations

Coordinate Systems and Parametric Equations. The parametric equations of a parabola with directrix x = -a and focus (a, 0) is:. y. M. P(at 2 , 2at). x. O. F(a, 0). A(-a, 0). EXAMPLE 1 A parabola has cartesian equation y 2 = 8x.

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Coordinate Systems and Parametric Equations

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  1. Coordinate Systems and Parametric Equations

  2. The parametric equations of a parabola with directrix x = -a and focus (a, 0) is: y M P(at2, 2at) x O F(a, 0) A(-a, 0)

  3. EXAMPLE 1 A parabola has cartesian equation y2 = 8x. Find the coordinates of the point on the parabola corresponding to the value t = 3/2 y2 = 4ax y2 = 8x So, a = 2 Parametrically:

  4. The parametric equations of a rectangular hyperbola can be expressed as:

  5. EXAMPLE 2 Show that any point , where t ≠ 0, lies on the rectangular hyperbola with cartesian equation xy = 9.

  6. EXAMPLE 3 The diagram shows part of a parabola C with focus F. Any point P On C has coordinates given by (t2, 2t) where t is any real number. Write down the coordinates of F. Point P(t2, 2t) on C is such that angle PFO = 45° Using the focus-directrix property of C, or otherwise, show that PF = 1 + t2 Find an expression, in terms of t, for OP2 and hence show that Find the exact area of triangle OPF, giving your answers in the form for integers p and q to be stated. y C P 45° O x F

  7. y (a) F(1, 0) x = -1 (b) By the focus-directrix property of a parabola, PF = PM C t2 1 + t2 PM = So, PF = 1 + t2 1 (t2, 2t) P M (c) P OP2 = ON2 + PN2 45° OP2 = (t2)2 + (2t)2 2t (-1, 0) O F (1, 0) x N OP2 = t4 + 4t2 O N t2

  8. y P x = -1 (c) 1 + t2 C 45° t2 1 1 F O (t2, 2t) P M OP2 = OF2 + PF2– 2(OF)(PF)CosF t4 + 4t2 = 1 + (1 + t2)2 – 2(1+t2) 45° (-1, 0) O F (1, 0) x N t4 + 4t2 = 1 + 1 + 2t2 + t4 – t2 2t2 + t2 = 2 –

  9. P (d) 2t 45° 1 F O Area = ½ base x height Area = ½ (1)(2t) = t But t = Area = p = q = -1

  10. EXAMPLE 4 The diagram shows part of a rectangular hyperbola, C. is any point on C. Show that the gradient of the line OP is is given by . Point Q on the line y = k – 2x, where k > 0 and is a constant, is such that the distance OQ is as short as possible. Given that Q is also on C, find the exact value of t corresponding to Q. Hence find the (i) coordinates of Q (ii) value of k. y O x

  11. (a) Gradient y O x

  12. (a) Gradient -2 Gradient of line = (b) y Gradient of OQ = (c) (i) Coordinates of Q are: = (ii) O x

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