Parabola

INTRODUCTION This chapter focusses on parabolic curves, which constitutes one category of various curves obtained by slicing a cone by a plane, called conic sections. A cone (not necessarily right circular) can be out in various ways by a plane, and thus different types of conic sections are obtained. Let us start with the definition of a conic section and then we will see how are they obtained by slicing a right circular cone. CONIC SECTIONS A conic section, or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line.  The fixed point is called the Focus.  The fixed straight line is called the Directrix.  The constant ratio is called the Eccentricity denoted by e. PS e PM   The line passing through the focus & perpendicular to the directrix is called the Axis. A point of intersection of a conic with its axis is called a Vertex. General Equation of a Conic If S is (p, q) & directrix is x + my + n = 0 then PS = PS = e  PM (x –  ) 2  (y – )2 & PM = | l x  my  n | l 2  m2 (2 + m2) [(x – p)2 + (y – q)2] = e2 (x + my + n)2 Which is of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 SECTION OF RIGHT CIRCULAR CONE BY DIFFERENT PLANES A right circular cone is as shown in the figure – 1 A axis Ge ner ato r Vertex B Circular Base Figure 1 PARABOLA (i) Section of a right circular cone by a plane passing through its vertex is a pair of straight lines passing through the vertex as shown in the figure - 2. V P Q Figure –2 (ii) Section of a right circular cone by a plane parallel to its base is a circle as shown in the figure  3. A B Figure- 3 (iii) Section of a right circular cone by a plane parallel to a generator of the cone is a parabola as shown in the figure–4. (iv) Section of a right circular cone by a plane neither parallel to any generator of the cone nor perpendicular or parallel to the axis of the cone is an ellipse or hyperbola as shown in the figure  5 & 6. Figure-4 A Figure -5 Figure -6 B 3D View Distinguishing Various Conics The nature of the conic section depends upon the position of the focus S w.r.t. the directrix & also upon the value of the eccentricity e. Two different cases arise. Case (I) When The Focus Lies On The Directrix. In this case   abc + 2fgh  af2  bg2  ch2 = 0 & the general equation of a conic represents a pair of straight lines if: e > 1  h2 > ab the lines will be real & distinct intersecting at S. e = 1  h = ab the lines will coincident. e < 1  h < ab the lines will be imaginary. 2 2 Case (II) When The Focus Does Not Lie On Directrix. a parabola an ellipse e = 1;   0, 0 < e < 1;   0; h² = ab h² < ab a hyperbola a re ctangular hyperbola e > 1;  0; e > 1;  0 h² > ab h² > ab; a + b = 0 PARABOLA PARABOLA A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed straight line (directrix). Standard equation of a parabola is y2 = 4ax. For this parabola : (i) Vertex is (0, 0) (ii) Focus is (a, 0) (iii) Axis is y = 0 (iv) Directrix is x + a = 0 Y M P L N Z A S X L' P' y2 = 4ax (a) Focal distance : The distance of a point on the parabola from the focus is called the focal distance of the point. (b) Focal chord : A chord of the parabola, which passes through the focus is called a focal chord. (c) Double ordinate : A chord of the parabola perpendicular to the axis of the symmetry is called a double ordinate. (d) Latus rectum : A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called the latus rectum. For y2 = 4ax.  Length of the latus rectum = 4a.  Length of the semi latus rectum = 2a.  Ends of the latus rectum are L(a, 2a ) & L'(a, – 2a ) (i) Perpendicular distance from focus on directrix = half the latus rectum. (ii) Vertex is middle point of the focus & the point of intersection of directrix & axis. (iii) Two parabolas are said to be equal if they have the same latus rectum. Ex. Find the equation of the parabola whose focus is at (– 1, – 2) and the directrix is x – 2y + 3 = 0. Sol. Let P(x, y) be any point on the parabola whose focus is S(– 1, – 2) and the directrix x – 2y + 3 = 0. Draw PM perpendicular to directrix x – 2y + 3 = 0. Then by definition, SP = PM  SP 2 = PM2   x  2y  3  (x + 1)2 + (y + 2)2 =    1 4   5 [(x + 1)2 + (y + 2)2] = (x – 2y + 3)2  5(x + y + 2x + 4y + 5) = (x + 4y + 9 – 4xy + 6x – 12y)  4x2 + y2 + 4xy + 4x + 32y + 16 = 0 2 2 M 2 x – 2y + 3 = 0 2 2 P(x, y) S(–1, –2) This is the equation of the required parabola. Ex. Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola 9y2 – 16x – 12y – 57 = 0. 2 Sol. 2 16  61 2 The given equation can be rewritten as  y     x   which is of the form Y = 4AX.   3 9 16  61 2  Hence the vertex is   ,   16 3  The axis is y – 2 2 = 0  y = 3 3 The directrix is X + A = 0  61 61 44 613 613 x +  00    xx 16 16 99 144 144 The focus is X = A and Y = 0  x  61 4 2  and y   0 16 9 3 4 85 2  focus =   ,  14 4 3  Length of the latus rectum = 4A = 16 9 The tangent at the vertex is X = 0  x 61 . 16 Ex. The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is - Sol. The length of latus rectum = 2 × perp. from focus to the directrix 2 2  4 (3 )  3 (1)  (4 ) 2 2  14 17 PARABOLA PARAMETRIC REPRESENTATION The simplest & the best form of representing the coordinates of a point on the parabola is (at², 2at) i.e. the equations x = at² & y = 2at together represents the parabola y² = 4ax, t being the parameter. Parametric form for : y2 = – 4ax (–at2, 2at) x2 = 4ay (2at , at2) x2 = – 4ay (2at , – at2) TYPE OF PARABOLA Four standard forms of the parabola are y2 = 4ax ; y2 = – 4ax ; x2 = 4ay ; x2 = –4ay Z Y Y M X' x=–a Y L P(x,y) T A N Y' X' X 0 S X P' Y' y2 = 4ax Parabola y=a (0,a) S Z' (–a,0) (a,0) S L' Z' Z Y X' 0 Z' y=–a Z' X' Z (0,–a) Y' y2 = –4ax Vertex Focus Axis Directrix X 0 Z X\ S Y' x2 = 4ay x2 = –4ay Length of Ends of Parametric Focal Latus rectum Latus rectum equation length 2 4a (a, ±2a) x +a (at ,2at) 2 4a (–a, ±2a) x–a (–at ,2at) 2 (0,0) (a,0) y= 0 x=–a 2 (0,0) (–a,0) y = 0 x=a 2 (0,0) (0,a) x = 0 y=–a 4a (±2a, a) (2at,at ) 2 (0,0) (0,–a) x = 0 y=a 4a (±2a, –a) (2at, –at ) y = 4ax y =–4ax x = +4ay x = –4ay 2 (y–k) = 4a(x–h) (h,k) 2 (x–p) = 4b(y–q) (p,q) (h+a,k) y = k x+a– h =0 4a (h+a, k±2a) (p, b+q) x=p y+b–q=0 4b (p±2a,q+a) 2 y+a 2 y–a 2 (h+at ,k+2at) x–h+a 2 (p+2at,q+at ) y–q+b Ex. The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola. Sol. Focus of the parabola is the mid-point of the latus rectum.  S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is y–4= 0 xx77    yy44 5 3 Length of the latus rectum = (5 – 3) = 2 Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabolas, one concaveigh towards and the other concave leftwards. The vertex of the first parabola is (6.5, 4) and its equation is (y – 4) 2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0). The equation of the second parabola is (y – 4)2 = –2(x – 7.5). It meets the x-axis at (–0.5, 0). Ex. Find the parametric equation of the parabola (x – 1)2 = –12 (y – 2) Sol.  4a = – 12  a = – 3, y – 2 = at2 x – 1 = 2 at  x = 1 – 6t, y = 2 – 3t2 POSITION OF A POINT WITH RESPECT TO A PARABOLA The point (x1 , y1) lies outside, on or inside the parabola y² = 4ax according as the expression y1²  4ax1 is positive, zero or negative. Outside Inside P(x1, y1) S1 : y12 – 4ax1 S1 < 0  Inside S1 > 0  Outside Ex. Check whether the point (3, 4) lies inside or outside the paabola y2 = 4x. Sol. y2 – 4x = 0  S1 y12 – 4x1 = 16 – 12 = 4 > 0  (3, 4) lies outside the parabola. Ex. Find the value of  for which the point ( – 1, ) lies inside the parabola y2 = 4x. Sol.  Point ( – 1, ) lies inside the parabola y2 = 4x  y12 –4ax1 < 0  2 – 4( – 1) < 0  2 – 4 + 4 < 0 ( – 2)2 < 0   LINE & A PARABOLA The line y = mx + c meets the parabola y² = 4ax in two points real, coincident or imaginary according as a  cm   condition of tangency is, c = a/m. Tangent Secant Length of the chord intercepted by the parabola A on the line y = m x + c is :  4  2  2  a(1  m )(a  mc) . m  B 1. The equation of a chord joining t1 & t2 is 2x  (t1 + t2) y + 2 at1 t2 = 0. 2. If t 1 & t 2 are the ends of a focal chord of the parabola y² = 4ax then t 1t 2 = 1. Hence the a 2a  coordinates at the extremities of a focal chord can be taken as (at², 2at) &  2 ,   t  t Focal chord A S (focus) B 3. Length of the focal chord making an angle  with the x axis is 4acosec²  PARABOLA Ex. Discuss the position of line y = x + 1 with respect to parabola y2 = 4x. Sol. Solving we get (x + 1) 2 = 4x  (x – 1)2 = 0 so y = x + 1 is tangent to the parabola. Ex. If the line y = 3x +  intersect the parabola y2 = 4x at two distinct points then set of values of  is - Sol. Putting value of y from the line in the parabola (3x + )2 = 4x  9x2 + (6 – 4)x + 2 = 0  line cuts the parabola at two distinct points  D>0  4(3 – 2)2 – 4.92 > 0  92 – 12 + 4 – 92 > 0   < 1/3 Hence,  (–, 1/3) Ex. If t1, t2 are end points of a focal chord then show that t1 t2 = –1. Sol. Let parabola is y2 = 4ax 2 since P, S & Q are collinear  mPQ = mPS  2 2t1 = 2 t1  t 2 t1  1  t12 – 1 = t12 + t1t2  t1t2 = – 1 P (at1 , 2at1) S(a, 0) 2 Q (at2 , 2at2) TANGENT TO THE PARABOLA y2 = 4ax (a) Point form Equation of tangent to the given parabola at its point (x1, y1) is yy1 = 2a (x + x1) (b) Slope form Equation of tangent to the given parabola whose slope is 'm', is y = mx + a , m  0  m 2a   a Point of contact is  2 ,  m  m (c) Parametric form Equation of tangent to the given parabola at its point P(t), is ty = x + at2 Point of intersection of the tangents at the point t1 & t2 is [ at1t2, a(t1+ t2)]. Equation of tangent at a point on the parabola can be obtained by replacement method or using derivatives. In replacement method, following changes are made to the second degree equation to obtain T. x2  x x1, y2  y y1, 2xy  xy1 + x1y, 2x  x + x1, 2y  y + y1 So, it follows that the targents are : (i) y y1 = 2 a (x + x1) at the point (x1, y1) ; (ii) y = mx + (iii) t y = x + a t² at (at², 2at). (iv) Point of intersection of the tangents at the point t1 & t2 is { at1 t2 , a(t1 + t2) }. a a 2a  (m  0) at  2 ,  m m m  Ex. Find the equation to the tangents to the parabola y2 = 9x which goes through the point (4, 10). Sol. Equation of tangent to parabola y2 = 9x is y = mx + 9 4m Since it passes through (4, 10)   9 4m 16 m2 – 40 m + 9 = 0 10 = 4m + m=  1 9 , 4 4 equation of tangent’s are y = & Ex. y= x +9 4 9 x + 1. 4 A tangent to the parabola y2 = 8x makes an angle of 45° with the straight line y = 3x + 5. Find its equation and its point of contact. Sol. Let the slope of the tangent be m 3 m 1  3m  tan45° =  1  3m   (3  m)  m = –2 or 1 2 As we know that equation of tangent of slope m to the parabola y 2 = 4ax is y = mx +  a 2a  point of contact is  2 , m  m 1  for m = –2, equation of tangent is y = –2x – 1 and point of contact is  ,  2  2  for m = 1 1 , equation of tangent is y = x + 4 and point of contact is (8, 8) 2 2 a and m PARABOLA Ex. Prove that the straight line y = mx + c touches the parabola y2 = 4a (x + a) if c = ma + Sol. Equation of tangent of slope ‘m’ to the parabola y2 = 4a(x + a) is a 1  y = m(x + a) +  y = mx + a  m   m m  but the given tangent is y = mx +c a  c = am + m a m Pair of Tangents The equation to the pair of tangents which can be drawn from any point (x1, y1) to the parabola y² = 4ax is given by: SS1 = T² where : S  y²  4ax ; S1 = y1²  4ax1 ; T  y y1  2a(x + x1). A (x1, y1 ) P B Ex. Write the equation of pair of tangents to the parabola y2 = 4x drawn from a point P(–1, 2) Sol. We know the equation of pair of tangents are given by SS 1 = T²  (y2 – 4x) (4 + 4) = (2y – 2 (x – 1))2  8y2 – 32x = 4y2 + 4x2 + 4 – 8xy + 8y – 8x  y2 – x2 + 2xy – 6x – 2y = 1 NORMALS TO THE PARABOLA y² = 4ax Normal is obtained using the slope of tangent. Slope of tangent at (x1 , y1) =  (a) 2a y1 y Slope of normal = – 1 2a Point form Equation of normal to the given parabola at its point (x1, y1) is y – y1 = – (b) y1 (x – x1) 2a Slope form Equation of normal to the given parabola whose slope is 'm', is y = mx – 2am – am3 foot of the normal is (am2, – 2am) (c) Parametric form Equation of normal to the given parabola at its point P(t), is y + tx = 2at + at3 P Normal (i) Point of intersection of normals at t1 & t2 is (a(t12 + t22 + t1t2 + 2), – at1t2 (t1 + t2)). (ii) If the normal to the parabola y2 = 4ax at the point t1, meets the parabola again at the point t2,  2 then t2 = –  t1 + . t 1   (iii) If the normals to the parabola y2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1t2 = 2 ; t3 = – (t1 + t2) and the line joining t1 & t2 passes through a fixed point (–2a, 0). (iv) If normal drawn to a parabola passes through a point P(h,k) then k = mh – 2 am – am 3 , i.e. am3 + m (2a – h) + k = 0. This gives m1 + m2 + m3 = 0 ; m1m2+m2m3+m3m1 = 2a  h k ; m1m2m3 = a a where m1, m2, & m3 are the slopes of the three concurrent normals :  Algebraic sum of slopes of the three concurrent normals is zero.  Algebraic sum of ordinates of the three co-normal points on the parabola is zero.  Centroid of the  formed by three co-normal points lies on the axis of parabola (x-axis). Ex. If the normal at point ‘t1’ intersects the parabola again at ‘t2’ then show that t2 = –t1 – Sol. Slope of normal at P = – t1 and slope of chord PQ =  – t1 = 2 t1  t 2  2 t2 = – t 1 – . t1 If two normals drawn from any point to the parabola y2 = 4ax make angle  and  with the axis such that tan . tan  = 2, then find the locus of this point. Sol. P (at , 2at ) Q (at , 2at ) 2 t1 + t 2 = – t1 Ex. 2 t1  t 2 2 t1 Let the point is (h, k). The equation of any normal to the parabola y2 = 4ax is y = mx – 2am – am3 passes through (h, k) k = mh – 2am – am3 am3 + m(2a – h) + k = 0 ....(i) m1, m2, m3 are roots of the equation, then m1. m2. m3 = – but m1m2 = 2, m3 = – k 2a m3 is root of (i) 3 k  k a   (2 a  h )  k  0  k2 = 4ah  2a  2a Thus locus is y2 = 4ax.  k a PARABOLA Ex. If the normals at points t1, t2 meet at the point t3 on the parabola then prove that (i) Sol. t1 t2 = 2 (ii) t1 + t 2 + t 3 = 0 Since normal at t1 & t2 meet the curve at t3   t3 = – t1 – 2 t1 .....(i) t3 = – t 2 – 2 t2 .....(ii) (t12 + 2) t2 = t1 (t22 + 2) t1t2 (t1 – t2) + 2 (t2 – t1) = 0  t1  t 2 , t 1t2 = 2 Hence (i) t1 t2 = 2 ......(iii) from equation (i) & (iii), we get t3 = – t1 – t2 Hence (ii) t1 + t2 + t3 = 0 LENGTH OF SUBTANGENT & SUBNORMAL 2 P (at, 2at) PT and PG are the tangent and normal respectively at the point P to the parabola y2 = 4ax. Then TN = length of subtangent = twice the abscissa of the point P (Subtangent is always bisected by the vertex) T N G NG = length of subnormal which is constant for all points on the parabola & equal to its semilatus rectum (2a). DIRECTOR CIRCLE Locus of the point of intersection of the perpendicular tangents to the parabola y2 = 4ax is called the director circle. It’s equation is x + a = 0 which is parabola’s own directrix. Ex. The circle drawn with variable chord x + ay – 5 = 0 (a being a parameter) of the parabola y2 = 20x as diameter will always touch the line - Sol. Clearly x + ay – 5 = 0 will always pass through the focus of y2 = 20x i.e. (5, 0). Thus the drawn circle will always touch the directrix of the parabola i.e. the line x + 5 = 0. CHORD JOINING TWO POINTS The equation of a chord of the parabola y2 = 4ax joining its two points P(t1) and Q(t2) is y(t1 + t2) = 2x + 2at1t2  If PQ is focal chord then t1t2 = –1.   a  2a  Extremities of focal chord can be taken as (at2, 2at) &  2 ,  t  t Ex. Through the vertex O of a parabola y2 = 4x chords OP and OQ are drawn at right angles to one another. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point. Sol. The given parabola is y2 = 4x ...... (i) Let P   t12 , 2 t1  , Q   t 22 , 2 t 2  Slope of OP = 2 2t1 2 and slope of OQ =  2 t t1 t1 2 4  1 or t1t2 = –4 t1 t 2 Since OP  OQ, ...... (ii) The equation of PQ is y(t1 + t2) = 2 (x + t1t2)   4 y  t1   = 2(x – 4) t1   [from (ii)]   4 2(x – 4) – y  t1   = 0 t1     variable line PQ passes through a fixed point which is point of intersection of L1= 0 & L2= 0 i.e. (4, 0) L1 + L2 = 0 CHORD OF CONTACT Equation to the chord of contact of tangents drawn from a point P(x1, y1) is yy1 = 2a (x + x1). A P (x , y ) Chord of contact B The area of the triangle formed by the tangents from the point (x1, y1) & the chord of contact is (S )3/2 1 (y1²  4ax1)3/2 = 1 2a 2a Ex. Find the length of chord of contact of the tangents drawn from point (x1, y1) to the parabola y2 = 4ax. Sol. Let tangent at P(t1) & Q(t2) meet at (x1, y1)  at1t2 = x1  PQ = =a & a(t1 + t2) = y1 (at12  at 22 ) 2  (2a(t1  t 2 ))2 ((t1  t 2 ) 2  4t1t 2 )((t1  t 2 ) 2  4) = (y12  4ax1 )(y12  4a 2 ) a2 PARABOLA Ex. If the line x – y – 1 = 0 intersect the parabola y2 = 8x at P & Q, then find the point of intersection of tangents at P & Q. Sol. Let (h, k) be point of intersection of tangents then chord of contact is yk = 4(x + h) 4x – yk + 4h = 0 ....... (i) But given line is x–y–1=0 ....... (ii) Comparing (i) and (ii)  4 k 4 h   1 1 1  point  (–1, 4)  h = – 1, k = 4 CHORD WITH A GIVEN MIDDLE POINT Equation of the chord of the parabola y2 = 4ax whose middle point is (x1, y1) is y – y1 = 2a  x  x1  . y1 M(x,y) This reduced to T = S1, where T  yy1 – 2a (x + x1) & S1  y12 – 4ax1. Ex. Find the locus of middle point of the chord of the parabola y2 = 4ax which pass through a given point (p, q). Sol. Let P(h, k) be the mid point of chord of parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah. Since it passes through (p, q)  qk – 2a (p + h) = k2 – 4ah  Required locus is y2 – 2ax – qy + 2ap = 0. Ex. Find the locus of middle point of the chord of the parabola y2 = 4ax whose slope is ‘m’. Sol. Let P(h, k) be the mid point of chord of parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah. but slope =  2a =m k locus is y = 2a m DIAMETER The locus of the middle points of a system of parallel chords of a Parabola is called a DIAMETER. Equation to the diameter of a parabola is y = 2a/m, where m = slope of parallel chords. IMPORTANT CONCEPT (i) If the t angent & nor mal at any point ‘P’ of the parabo la int ers ect the axis at T & G then ST = SG = SP where ‘S’ is the focus. In other words the tangent and the normal at a point P on the parabola are the bisectors of the angle between the focal radius SP & the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to the axis of the parabola after reflection. y y P M R T O S x G O S x Reflection property Directrix (ii) The portion of a tangent to a parabola cut off between the directrix & the curve subtends a right angle at the focus. See figure above. (iii) The tangents at the extremities of a focal chord intersect at right angles on the directrix, and hence a circle on any focal chord as diameter touches the directrix. Also a circle on any focal radii of a point P (at2, 2at) as diameter touches the tangent at the vertex and intercepts a chord of length a 1  t 2 on a normal at the point P. (iv) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent at the vertex. (v) If the tangents at P and Q meet in T, then:  TP and TQ subtend equal angles at the focus S.  ST 2 = SP. SQ &  The triangles SPT and STQ are similar. P T S Q TSP = TSQ PARABOLA (vi) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of any focal chord of the 2bc 1 1 1 i.e. .   b+c b c a parabola is ; 2a = (vii) The area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. (viii) If normal are drawn from a point P(h, k) to the parabola y2 = 4ax then k = mh  2am  am3 i.e. am3 + m(2a  h) + k = 0. 2a  h ; m1 m2 m3 =  k . a a Where m1, m2, & m3 are the slopes of the three concurrent normals. Note that m1 + m2 + m3 = 0 ; m1m2 + m2m3 + m3m1 = A P(h, k) A, B, C  Conormal points B C  algebraic sum of the slopes of the three concurrent normals is zero.  algebraic sum of the ordinates of the three conormal points on the parabola is zero  Centroid of the  formed by three conormal points lies on the xaxis.  Condition for three real and distinct normals to be drawn froma point P (h, k) is h > 2a & k2 < (ix) 4 (h – 2a)3. 27A Length of subtangent at any point P(x, y) on the parabola y² = 4ax equals twice the abscissa of the point P. Note that the subtangent is bisected at the vertex. y P T O D N x TD = 2(OD) , DN = 2a (x) Length of subnormal is constant for all points on the parabola & is equal to the semi latus rectum. See figure above. (xi) Tangents and Normals at the extremities of the latus rectum of a parabola L N (–a,0) S L' 2 (a,2a) G (3a,0) (a,–2a) y = 4ax constitute a square, their points of intersection being (–a, 0) & (3a, 0). (xii) The circle circumscribing the triangle formed by any three tangents to a parabola passes through the focus. (xiii) If a family of straight lines can be represented by an equation 2P + Q + R = 0 where  is a parameter and P, Q, R are linear functions of x and y then the family of lines will be tangent to the curve Q2 = 4PR. (a) The two tangents at the extremities of focal chord meet on the foot of the directrix. (b) Figure LNL'G is square of side 2 2 a PARABOLA 1. Conic Section A conic section, or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line. 2. (a) The fixed point is called the FOCUS. (b) The fix straight line is called the DIRECTRIX. (c) The constant ratio is called the ECCENTRICITY denoted by e. (d) The line passing through the focus & perpendicular to the directrix is called the AXIS. (e) A point of intersection of a conic with its axis called a VERTEX. General Equation of a Conic : Focal Direction Property The general equation of a conic section with focus(p, q) & directrix lx + my + n = 0 is (l2 + m2 ) [(x – p)2 + (y-q2)] = e2 (lx + my + n)2  ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 3. Distinguishing Between the Conic The nature of the conic section depends upon the position of the focus S w.r.t. the directrix & also upon the value of the eccentricity e. Two different cases arise. Case (i) When the Focus Lies on the Directrix : In this case D  abc + 2 fgh – af2 – bg2 – ch2 = 0 & the general equation of a conic represents a pair of straight lines and if : e > 1, h2 > ab the lines will be real & distinct intersecting at S. e = 1, h2 = ab the lines will coincident. e < 1, h2 < ab the lines will be imaginary. Case (ii) When the Focus does not Lie on the Directrix : The conic represents : a parabola an ellips e a hyperbola a rectang ular hyperbola e = 1; D  0 01 e>1; D 0 2 h = ab 4. 2 h < ab 2 h > ab 2 h > ab ; a + b = 0 Parabola A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed straight line (directrix). Standard equation of a parabola is y2 = 4 ax. For this parabola : (i) Vertex is (0,0) (ii) Focus is (a, 0) (a) (iii) Axis is y = 0 (iv) Directrix is x + a = 0 Focal Distance The distance of a point on the parabola from the focus is called the FOCAL DISTANCE OF THE POINT. (b) Focal Chord : A chord of the parabola, which passes through the focus is called a FOCAL CHORD. (c) Double Ordinate : A chord of the parabola perpendicular to the axis of the symmetry is called a DOUBLE ORDINATE with respect to axis as diameter. (d) Latus Rectum : A focal chord perpendicular to the axis of parabola is called the LATUS RECTUM. For y2 = 4ax (i) Length of the latus rectum = 4a (ii) Length of the semi latus rectum = 2a (iii) Ends of the latus rectum are L(a, 2a) & L’(a, –2a) Note that 5. (i) Perpendicular distance from focus on directrix = half the latus rectum. (ii) Vertex is middle point of the focus & the point of intersection of directrix & axis. (iii) Two parabolas are said to be equal if they have latus rectum of same length. Parametric Representation The simplest & the best form of representing the co-ordinates of a point on the parabola y2 = 4ax is (at2, 2at). The equation x = at2 & y = 2at together represents the parabola y2 = 4ax, t being the parameter. 6. Type of Parabola Four standards forms of the parabola are y2 = 4ax ; y2 = –4ax ; x2 = 4ay ; x2 = –4ay M Z Y Y L X’ T (a,0) S A Z’ X N L’ x = -a P(x,y) Y’ X’ (-a,0) S x=a P’ Y’ Y y=a (0,a) S Z’ X 0 0 Y’ x2 = 4ay Z X X’ (0,-a) S y = -a Z’ y2 = - 4ax Y Z’ X 0 y2 = 4ax X’ Z Z Y’ x2 = - 4ay PARABOLA Parabola 2 y = 4ax 2 y = -4ax 2 x = +4ay 2 x = -4ay 2 (y-k) = 4a(x-h) 2 (x-p) = 4b (y-q ) 7. Length of Axis Directrix Latus rectum Ends of Latus rectum Parametric equation Vertex Focus (0,0) (a,0) y=0 x = -a 4a (a,± 2a) (at , 2at) (0,0) (-a,0) y=0 x= a 4a (-a,± 2a) (-at , 2at) (0,0) (0,0) (0,a) x=0 (0,-a) x=0 y = -a y=a 2 x-a 2 (2at, at ) y+a 2 (± 2a, -a) 4a x+a 2 (± 2a,a) 4a Focal length (2at, -at ) y-a 2 (h ,k) (h +a,k) y=k x+a-h = 0 4a (h+a, k ± 2a) (h +at , k+2at) x-h+a (p ,q ) (p , b +q) x = p y+b-q = 0 4b (p ± 2a, q+a) (q+2at, q +at ) y-q+b 2 Position of a Point Relative to a Parabola The point (x1 ; y1) lies outside, on or inside the parabola y2 = 4ax according as the expression y12 = –4ax1 is positive, zero ot negative. 8. Chord Joining two Points The equation of a chord of the parabola y2 = 4ax joining its two points P(t1) and Q(t2) is y(t1 + t2) = 2x + 2at1t2 Note 9. (i) If PQ is focal chord then t1t2 = –1. (ii)  a –2a  Extremities of focal chord can be taken as (at2, 2at) &  2 ,  t  t (iii) If t1t2 = k then chord always passes a fixed point (-ka, 0). Line & a Parabola (a) The line y = mx + c meets the parabola y2 = 4ax in two points real, coincident or imaginary according as a > =