Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Mathematics

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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**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)**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**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**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?**Length of Focal Chord**Length of focal chord joining ‘t1’, ‘t2’ on y2 = 4ax AB = length of focal chord = SA + SB**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**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:**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**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:**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**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:**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)**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)**Point of Intersection of Two Normals at y2 = 4ax**Parametric form Normals at ‘t1’, ‘t2’ be Solving these we get**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**For Other Standard Parabolas Tangent in Parametric(‘t’)**Form**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**For Other Standard Parabolas Normals in Parametric(‘t’)**Form**Other Method for Finding Tangents/Normals**Let Curve be y = f(x) Tangent at (x1,y1) Normal at (x1,y1)**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**Condition of tangency ofy = mx + c to y2 = 4ax is**Solution Hence, answer is (a).**Show that x = my + c touches**the parabola Class Exercise - 2**Substituting x, we get**if line touches D = 0 Solution**Class Exercise - 3**Find the locus of the point ofintersection of tangents at theextremities of a focal chord ofy2 = 4ax.**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**Class Exercise - 4**If x + y = k is normal to y2 = 12xthen find k and the foot of the normal.**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.**Class Exercise - 5**Prove that in a parabola semi-latusrectum is the harmonic mean of thesegments of a focal chord.**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**H.M. of these two is**= semi latus rectum Solution contd.. Hence proved.**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**Equation of circle is**Solution**If is imaginary**Real points of intersections are Solution contd.. For point of intersection with y2 = 2px Hence answer is (a), (b).**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.**Let A be slope of normal atA is –t1. Let normal again**cuts theparabola y2 = 4ax at Solution**Slope of AB is**Solution contd.. From (i) and (ii)**Class Exercise - 8**Find the locus of the point ofintersection of those normals tothe parabola y2 = 8x which areat right angles to each other.**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**as [from (ii)]**[from (i)] Solution contd.. Solving we get**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**Solution**Focus (a, 0) Moving point be (h, k) such that k2 = 4ah To find locus of mid point Hence, answer is (c).**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.