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Mathematics

Mathematics. Session. Parabola Session 2. Session Objective. 1. Position of a point with respect to a parabola 2. Parametric form of parabola 3. Focal chord 4. Intersection of a line and a parabola 5. Tangent in various forms 6. Normal in various forms 7. Other Standard Parabolas.

hoshiko-miyamoto
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Mathematics

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  1. Mathematics

  2. Session Parabola Session 2

  3. Session Objective • 1. Position of a point with respect to a parabola • 2. Parametric form of parabola • 3. Focal chord • 4. Intersection of a line and a parabola • 5. Tangent in various forms • 6. Normal in various forms • 7. Other Standard Parabolas

  4. Position of a Point With Respect to a Parabola • The point (h, k) lies outside, on, or inside the parabola y2 = 4ax according as k2 – 4ah > , =, < 0 • Denote S  y2 – 4ax and S1 k2 – 4ah • then point is out, on or in as S1 >, =, < 0 • S1 is obtained by substituting point in the equation of curve S = 0, (S1 is known as power of the point)

  5. Parametric Form of Parabola y2 = 4ax x = at2, y = 2at where t is parameter General point on y2 = 4 ax can be written as P(at2, 2at) called ‘t’ point Parametric form of (y – k)2 = 4a(x – h) is obtained as x – h = at2 , y – k = 2at i.e. x = h + at2 , y = k + 2at

  6. Equation of Chord Joining Any Two Points on the Parabola Chord joining ‘t1’ , ‘t2’ lying on y2 = 4ax is given by Parametric Form Chord joining (x1,y1), (x2,y2) lying on y2 = 4ax is given by Point Form

  7. or or Condition of Focal chord and Focal distance Chord joining 2 points Condition for focal chord (chord passing through focus) As above lines passes through (a,0) Focal Distance of (x1,y1) on y2 = 4 ax (distance from focus) Why?

  8. Length of Focal Chord Length of focal chord joining ‘t1’, ‘t2’ on y2 = 4ax AB = length of focal chord = SA + SB

  9. Intersection of a Line and a Parabola Let the line be y = mx + c and parabola be y2 = 4ax then at point of intersection Nature of Intersection • Real & distinct or Line cut parabola in 2 pts : c < a/m • Real & Equal or Line Touches parabola: c = a/m • Imaginary or Line can not intersect parabola: c > a/m

  10. Equation of Tangent in Parametric(‘t’) Form Parametric Form x = at2 , y = 2at Equation of Tangent at ‘t1’ i.e. (at12,2at1) Chord joining ‘t1’, ‘t2’ is (t1 + t2)y = 2x + 2at1t2 let t2 t1 where point of contact is (at12,2at1) and slope = 1/t1 Alternative using Differentiation:

  11. Equation of Tangent in Point(x1,y1) Form Point Form: Tangent at (x1,y1) on y2 = 4ax Chord joining 2 points is (y1 + y2)y = 4ax + y1y2 let y2 y1 T = 0 where point of contact is (x1,y1) and slope = 2a/y1 Working Rule for Finding T = 0 Replace x2 xx1 , y2  yy1 , x  (x+x1)/2 y  (y+y1)/2 , xy  (xy1+x1y)/2 in the equation of curve

  12. Equation of Tangent in Slope(m) Form Slope Form: Tangent of slope m to y2 = 4ax y = mx + c intersect y2 = 4ax in two real & equal points if c = a/m (done earlier) is the equation of tangent of slope m point of contact given by:

  13. Point of Intersection of Two Tangents at y2 = 4ax and Angle between them Parametric form Tangents at ‘t1’, ‘t2’ be Solving these we get x = at1t2 , y = a(t1 + t2) Angle between these tangents is given by

  14. Equation of Normal in Parametric(‘t’) Form Parametric Form x = at2 , y = 2at Equation of Normal at ‘t1’ i.e. (at12,2at1) Slope of tangent at ‘t1’ is 1/t1 slope of normal = –t1  Equation is y – 2at1 = –t1(x – at12) with Foot of normal = ‘t1’ Alternative using Differentiation:

  15. Equation of Normal in Point(x1,y1) Form Point Form: Normal at (x1,y1) on y2 = 4ax Slope of tangent = 2a/y1  slope of normal = –y1/2a with foot of normal at (x1,y1)

  16. Equation of Normal in Slope(m) Form Slope Form: Normal of slope m to y2 = 4ax Equation of Normal at ‘t1’ i.e. (at12,2at1) As slope = m  –t1 = m or t1 = –m is the equation of normal in slope form with foot of normal at (am2,–2am)

  17. Point of Intersection of Two Normals at y2 = 4ax Parametric form Normals at ‘t1’, ‘t2’ be Solving these we get

  18. S Out On In S 1 2 2 S> 0 S< 0 y = 4ax k – 4ah S= 0 1 1 1 2 2 S> 0 y = – 4ax k + 4ah S= 0 S< 0 1 1 1 2 2 S> 0 x= 4ay h – 4ak S= 0 1 1 S< 0 1 2 2 x= – 4ay h + 4ak S> 0 S< 0 S= 0 1 1 1 For Other Standard Parabolas Position of a Point

  19. For Other Standard Parabolas Parametric Forms

  20. For Other Standard Parabolas Tangent in Point(x1,y1) Form

  21. For Other Standard Parabolas Tangent in Parametric(‘t’) Form

  22. For Other Standard Parabolas Tangent in Slope(m) Form

  23. Equation of parabola Normal at (x1, y1) y2 = 4ax y2 = –4ax x2 = 4ay x2 = –4ay For Other Standard Parabolas Normals in Point(x1,y1) Form

  24. For Other Standard Parabolas Normals in Parametric(‘t’) Form

  25. For Other Standard Parabolas Normals in Slope(m) Form

  26. Other Method for Finding Tangents/Normals Let Curve be y = f(x) Tangent at (x1,y1) Normal at (x1,y1)

  27. Class Exercise

  28. Class Exercise - 1 The line y = mx + 1 is a tangent tothe parabola y2 = 4x if (a) m = 1 (b) m = 2(c) m = 4 (d) m = 3

  29. Condition of tangency ofy = mx + c to y2 = 4ax is Solution Hence, answer is (a).

  30. Show that x = my + c touches the parabola Class Exercise - 2

  31. Substituting x, we get if line touches D = 0 Solution

  32. Class Exercise - 3 Find the locus of the point ofintersection of tangents at theextremities of a focal chord ofy2 = 4ax.

  33. Let be the extremities of the focal chord thent1t2 = –2. Point of intersection oftangents is given by x = –a is the equation of directrix of the parabola Required locus is x = –a Solution

  34. Class Exercise - 4 If x + y = k is normal to y2 = 12xthen find k and the foot of the normal.

  35. y = –x + k, equation of normal in slopeformisy = mx – 2am – am3 with foot of normal as Solution = 6 + 3 = 9 and foot of normal is (3, 6) Note: Since, we know slope of normal, we haveused slope form.

  36. Class Exercise - 5 Prove that in a parabola semi-latusrectum is the harmonic mean of thesegments of a focal chord.

  37. Let chord joining ‘t1’, ‘t2’ be the focal chord points are Segments of the focal chord are Solution Let the equation of parabola be y2 = 4ax Semi latus rectum = 2a

  38. H.M. of these two is = semi latus rectum Solution contd.. Hence proved.

  39. Consider a circle with its centre lyingon the focus of the parabola y2 = 2pxsuch that it touches the directrix of theparabola. Then a point of intersectionof the circle and the parabola is/are • (b) • (c) (d) Class Exercise - 6

  40. Equation of circle is Solution

  41. If is imaginary Real points of intersections are Solution contd.. For point of intersection with y2 = 2px Hence answer is (a), (b).

  42. Class Exercise - 7 A is the point on the parabolay2 = 4ax. The normal at A cuts theparabola again at the point B.If AB subtends a right angle atthe vertex of the parabola, findthe slope of AB.

  43. Let A be slope of normal atA is –t1. Let normal again cuts theparabola y2 = 4ax at Solution

  44. Slope of AB is Solution contd.. From (i) and (ii)

  45. Class Exercise - 8 Find the locus of the point ofintersection of those normals tothe parabola y2 = 8x which areat right angles to each other.

  46. Let the normals be drawn at the points their slopes are Normals are given by Solution As normals are perpendicular t1.t2=–1 Solving we get

  47. as [from (ii)] [from (i)] Solution contd.. Solving we get

  48. The locus of the mid point of the linesegment joining the focus to a movingpoint on the parabola y2 = 4ax is anotherparabola with directrix • x = –a(b) • (c) x = 0 (d) Class Exercise - 9

  49. Solution Focus (a, 0) Moving point be (h, k) such that k2 = 4ah To find locus of mid point Hence, answer is (c).

  50. Class Exercise - 10 Show that the locus of a point thatdivides a chord of slope 2 of theparabola y2 = 4ax internally in theratio 1 : 2 is a parabola. Find thevertex of this parabola.

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