Ellipse Notes

DEFINITIONS It is locus of a point which moves in such a way that the ratio of its distance from a fixed point and a fixed line (not passes through fixed point and all points and line lies in same plane) is constant (e), which is less than one. The fixed point is called - focus The fixed line is called -directrix. The constant ratio is called - eccentricity, it is denoted by 'e'. GENERAL EQUATION OF AN ELLIPSE Let (a, b) be the focus S, and lx + my + n = 0 is the equation of directrix. P'(x,y) Ex. Sol.  (x – a)2 + (y – b)2 = e2  SP = e PM (e is the eccentricity)  (l2 + m2) {(x – a)2 + (y – b)2} = e2{lx + my + n}2 (l x  my  n )2 (l 2  m ) axis S(a,b) M lx + my + n = 0 Let P(x, y) be any point on the ellipse. Then by definition. Find the equation to the ellipse whose focus is the point (– 1, 1), whose directrix is the straight 1 line x – y + 3 = 0 and eccentricity is . 2 Let P  (h, k) be moving point, P (h,k) M PS 1 e= = PM 2 2 S (–1, 1) 1  h k 3 2 2  (h + 1) + (k – 1) =   4  2   locus of P(h, k) is 8 {x2 + y2 + 2x – 2y + 2} = (x2 + y2 – 2xy + 6x – 6y + 9) 7x2 + 7y2 + 2xy + 10x – 10 y + 7 = 0. STANDARD EQUATION & DEFINITION Standard equation of an ellipse referred to its principal axes along the co-ordinate axes is b & b2 = a2 (1 – e2 )  a2 – b2 = a2 e2 . where e = eccentricity (0 < e < 1). FOCI : S  (ae, 0) & S'  (– ae, 0 ). (a) Equation of directrices : x= (b) a a & x . e e Vertices : A'  (–a, 0) & A  (a, 0). x 2 y2 + = 1 . where a > a 2 b2 ELLIPSE (c) (d) Major axis : The line segment A' A in which the foci S' & S lie is of length 2a & is called the major axis (a > b) of the ellipse. Point of intersection of major axis with directrix is called the foot of the  a  directrix (z)   , 0  .  e  Minor Axis : The y-axis intersects the ellipse in the points B’  (0,- b) & B  (0, b). The line segment B’B of length 2b (b < a) is called the Minor Axis of the ellipse. (e) Principal Axes : The major & minor axis together are called Principal Axes of the ellipse. (f) (g) Centre : The point which bisects every chord of the conic drawn through it is called the centre of the conic. C  (0,0) x2 y2 the origin is the centre of the ellipse 2  2  1 . a b Diameter : A chord of the conic which passes through the centre is called a diameter of the conic. (h) Focal Chord : A chord which passes through a focus is called a focal chord. (i) Double Ordinate : A chord perpendicular to the major axis is called a double ordinate. (j) Latus Rectum : The focal chord perpendicular to the major axis is called the latus rectum. 2b 2 (minor axis) 2   2a(1  e 2 ) a major axis (i) Length of latus rectum (LL') = (ii) Equation of latus rectum : x = ± ae. (iii)     b2  b2  b2  b2  Ends of the latus rectum are L  ae,  , L'  ae,  .  , L 1   ae,  and L 1 '   ae,  a  a  a  a      (k) Focal radii : SP = a –ex & S'P = a + ex  (l) Eccentricity : e  1– SP + S 'P = 2a = Major axis. b2 a2 (i) The sum of the focal distances of any point on the ellipse is equal to the major Axis. Hence distance of focus from the extremity of a minor axis is equal to semi major axis. i.e BS = CA. (ii) If the equation of the ellipse is given as x2 y2   1 & nothing is mentioned, then the rule is to a 2 b2 assume that a > b. Ex. Sol. If LR of an ellipse is half of its minor axis, then its eccentricity is 2 b2 b  As given 2b = a a   4a2(1 – e2) = a2 1 – e2 = 1/4  e=  4b2 = a2 3/2 Ex. Find the equation of the ellipse whose foci are (2, 3), (–2, 3) and whose semi minor axis is of length Sol. Here S is (2, 3) & S' is (–2, 3) and b = but 2 2 2 5 2 b = a (1 – e )  5 = a – 4  SS' = 4 = 2ae  a = 3. Hence the equation to major axis is y = 3 Centre of ellipse is midpoint of SS' i.e. (0, 3)  x2 (y  3 )2 x 2 (y  3 )2  1 or  1 Equation to ellipse is 2  2 a b 9 5  ae = 2 5 . Ex. Sol. 1 3 Since both focus lies on x-axis, therefore x-axis is major axis and mid point of focii is origin which is centre and a line perpendicular to major axis and passes through centre is minor axis which is y-axis. x2 y2 Let equation of ellipse is + =1 a2 b2 1  ae = 4 and e = (Given) 3  a = 12 and b 2 = a2 (1 – e2) Find the equation of the ellipse whose focii are (4, 0) and (– 4, 0) and eccentricity is   1 b2 = 144 1    9 b 2 = 16 × 8 b=8 2 Equation of ellipse is Ex. Sol. x2 y2 + =1 144 128 If minor-axis of ellipse subtend a right angle at its focus then find the eccentricity of ellipse. x2 y2 Let the equation of ellipse is 2 + 2 = 1 (a > b) a b   BSB = 2 and OB = OB  BSO =  OS = OB  e2 =  4 b2 = 1 – e2 a2 Another form of ellipse  ae = b  e= 1 2 x2 y2 + 2 = 1 , (a b As given 2b = 2ae  b = ae ...... (i)  b2 = 5a ...... (ii)    b 2 = a2 – b 2 2b2 = a2 a2 = 100, b2 = 50 [From (i)] ...... (iii) 2 Also Now since 2b  10 a b 2 = a 2 – a 2e 2 (ii), (iii) Hence equation of the ellipse will be x2 y2   1  x2 + 2y2 = 100 1 0 0 50 Similarly when a < b then required ellipse is 2x2 + y2 = 100 POSITION OF A POINT W.R.T. AN ELLIPSE The point P(x1, y1) lies outside, inside or on the ellipse according as ; Ex. Check whether the point P(3, 2) lies inside or outside of the ellipse Sol. S1   x12 y12   1 > < or = 0. a2 b2 x2 y2 + = 1. 25 16 9 4 9 1 + –1= + –1<0 25 16 25 4 Point P  (3, 2) lies inside the ellipse. AUXILIARY CIRCLE / ECCENTRIC ANGLE A circle described on major axis of ellipse as diameter is called the auxiliary circle. Let Q be a point on the auxiliary circle x² + y² = a² such that line through Q perpendicular to the x  axis on the way intersects the ellipse at P, then P & Q are called as the Corresponding Points on the ellipse & the auxiliary circle respectively. ‘’ is called the Eccentric Angle of the point P on the ellipse ( < ). Q  (a cos , a sin) P  (a cos , b sin)   l (P N) b Semi minor axis   l (Q N) a Semi major axis If from each point of a circle perpendiculars are drawn upon a fixed diameter then the locus of the points dividing these perpendiculars in a given ratio is an ellipse of which the given circle is the auxiliary circle. PARAMETRIC REPRESENTATION 2 y2 The equations x = a cos  & y = b sin  together represent the ellipse x a 2  2 = 1. b Where  is a parameter. Note that if P()  (a cos  b sin ) is on the ellipse then; Q()  (a cos  a sin ) is on the auxiliary circle. The equation to the chord of the ellipse joining two points with eccentric angles  &  is given by x  y   cos  sin  cos . a 2 b 2 2 Ex. Sol. Write the equation of chord of an ellipse   5  x2 y2 + =1 joining two points P   and Q   . 4    4  25 16   5    5    5           x y 4 4 4 4     4 4  cos + . sin = cos 5 4 2 2 2 Equation of chord is  3   3  x y . cos   + . sin   = 0 4 5 4    4  – x y + =0 5 4  4x = 5y Ex. Find the focal distance of a point P() on the ellipse Sol. Let 'e' be the eccentricity of ellipse.  x2 y2 + =1 a2 b2 (a > b) PS = e . PM a   a cos   e   =e  and PS = (a – a e cos) PS = e. PM   = e  a cos   a  e PS = a + ae cos  focal distance are (a ± ae cos) Note : PS + PS = 2a PS + PS = AA Ex. Find the angle between two diameters of the ellipse eccentricity angle  and  =  + Sol. Let ellipse is  . 2 x2 y2 =1 2 + a b2 Slope of OP = m1 = bsin  b = tan a cos  a Slope of OQ = m2 = bsin  b =– cot a cos  a  x2 y2 + 2 = 1. Whose extremities have 2 a b m1  m 2 tan = 1  m1m 2 = given  =  + b (tan   cot ) a b2 1 2 a =  2 2ab (a 2  b 2 )sin 2 ELLIPSE LINE AND AN ELLIPSE The line y = mx + c meets the ellipse is < = or > a²m² + b². x2 y2 = 1 in two points real, coincident or imaginary according as c² 2 + a b2 y x Hence y = mx + c is tangent to the ellipse a 2  2 = 1 if c² = a²m² + b². b 2 2 Ex. For what value of  does the line y = x +  touches the ellipse 9x2 + 16y2 = 144. Sol. Equation of ellipse is x2 y 2  1 16 9 9x2 + 16y2 = 144 or x2 y2 Comparing this with 2  2  1 then we get a2 = 16 and b2 = 9 a b and comparing the line y = x +  with y = mx + c  m = 1 and c =  If the line y = x +  touches the ellipse 9x2 + 16y2 = 144, then c2 = a2m2 + b2 Ex.  2 = 16 × 12 + 9  2 = 25  =±5 Find the set of value(s) of '' for which the line 3x – 4y +  = 0 intersect the ellipse points. Sol. Solving given line with ellipse, we get x2 y2 + = 1 at two distinct 16 9 (4y  ) 2 y2 + =1 9  16 9 y 2y 2 2 – + –1 = 0 18 9 144 Since, line intersect the parabola at two distinct points,  roots of above equation are real & distinct  D0   2 2 8    1  >0 – . (18)2 9  144  –12 2 <  < 12 2 TANGENT TO THE ELLIPSE x2 y2 + =1 : a 2 b2 xx1 yy1  2 1 a2 b 2 2 Note : For general ellipse replace x by (xx1), y by (yy1), 2x by (x + x1), 2y by (y + y1), 2xy by (xy1 + yx1) & c by (c). (a) Point form : Equation of tangent to the given ellipse at its point (x1, y1) is (b) Slope form : Equation of tangent to the given ellipse whose slope is 'm', is y = mx ± a 2 m 2  b 2   a2m mb 2 , Point of contact are  2 2 2 a 2 m2  b2  a m b    Note that there are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction. (c) Parametric form : Equation of tangent to the given ellipse at its point (a cos  , b sin  ), is xcosθ ysin θ + = 1 a b (i) There are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.These tangents touches the ellipse at extremities of a diameter.  cos (ii) Point of intersection of the tangents at the point  &  is,  a  cos  2   2 ,b sin  2    cos  2   (iii) The eccentric angles of the points of contact of two parallel tangents differ by . Ex. Find the equations of the tangents to the ellipse 3x2 + 4y2 = 12 which are perpendicular to the line y + 2x = 4. Sol. Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4. 1 2  mx – 2 = –1  m= Since 3x2 + 4y2 = 12 or x 2 y2  1 4 3 x2 y2  1 a 2 b2 a2 = 4 and b2 = 3 Comparing this with  So the equation of the tangent are y   y= 1 x±2 2 or 1 1 x 4 3 2 4 x – 2y ± 4 = 0. Ex. Find the lo cus of foot of perpend icular dr awn from centre to an y tange nt to the ellipse Sol. x2 y2 + = 1. a2 b2 Let P(h, k) be the foot of perpendicular to a tangent y = mx + a 2 m 2  b2 .......(i) from centre  k .m=–1 h  P(h, k) lies on tangent  k = mh +  m=– a 2 m 2  b2 from equation (ii) & (iii), we get 2  h2  a 2h 2 + b2 k   = k  k2   locus is (x2 + y2)2 = a2x2 + b2y2 h k .......(ii) , k) P(h .......(iii) ELLIPSE PAIR OF TANGENTS If P(x1,y1) be any point lies outside the ellipse x2 y2 + 2 =1, 2 a b and a pair of tangents PA, PB can be drawn to it from P. Then the equation of pair of tangents of PA and PB is SS1= T2 Ex. x12 y12 xx yy  2  1 , T  21  21  1 2 a b a b where S1  i.e. 2  x2 y2   x 12 y 12   xx yy   2  2  1   2  2  1  =  2 1  21  1  b b b  a  a   a How many real tangents can be drawn from the point (4, 3) to the ellipse x2 y2 + = 1. Find 16 9 the equation of these tangents & angle between them. Sol. Given point P  (4, 3) ellipse S  x2 y2 + –1=0 16 9 16 9 + –1=1>0 16 9  S1   Point P  (4, 3) lies outside the ellipse.  Two tangents can be drawn from the point P(4, 3). Equation of pair of tangents is SS1 = T 2   x 2 y2   1 . 1 =    16 9   xy x 2y x2 y2 x2 y2 + –1= + +1+ – – 6 2 3 16 9 16 9  – xy + 3x + 4y – 12 = 0  x=4&y=3 and angle between them = NORMAL TO THE ELLIPSE  4x 3y    1   16 9  2  (4 – x) (y – 3) = 0  2 x 2 y2  1 : a 2 b2 a 2 x b2 y – = a 2 – b 2 = a 2e 2. x1 y1 (a) Point form : Equation of the normal to the given ellipse at (x1 , y1 ) is (b) Slope form : Equation of a normal to the given ellipse whose slope is ‘m’ is y = mx m (c) Parametric form : Equation of the normal to the given ellipse at the point (acos, bsin) is axsec– bycosec= (a2 – b2). (a 2 – b 2 )m a 2 + b2m 2 . Ex. x 2 y2 Find the condition that the line x + my = n may be a normal to the ellipse 2  2  1. a b Sol. Equation of normal to the given ellipse at (a cos , b sin ) is ax by   a 2  b2 cos  sin  ......(i) If the line x + my = n is also normal to the ellipse then there must be a value of  for which line (i) and line x + my = n are identical. For that value of  we have l m n   2  a   b  (a  b2 )  cos     sin       and sin  =  bn m(a 2  b2 ) or an l (a 2  b2 ) ......(iii) ...... (iv) Squaring and adding (iii) and (iv), we get 1  Ex. cos   n2 (a  b2 )2 2 P and Q are corresponding points on the ellipse  a 2 b2   l 2  m 2  which is the required condition. x2 y2 = 1 and the auxiliary circles respectively. The 2 + a b2 normal at P to the ellipse meets CQ in R, where C is the centre of the ellipse. Prove that CR = a + b Sol. Let P  (acos , b sin)  Q  (a cos, a sin) Q Equation of normal at P is P (a sec) x – (b cosec ) y = a2 – b 2 ..........(i) equation of CQ is y = tan . x .........(ii) C Solving equation (i) & (ii), we get (a – b) x = (a2 – b 2) cos x = (a + b) cos, & y = (a + b) sin  R  ((a + b) cos, (a + b) sin)  CR = a + b x 2 y2  =1 16 9 Ex. Find the shortest distance between the line x + y = 10 and the ellipse Sol. Shortest distance occurs between two non-intersecting curve always along common normal. Let 'P' be a point on ellipse and Q is a point on given line for which PQ is common normal.  Tangent at 'P' is parallel to given line  Equation of tangent parallel to given line is (y = mx ± a 2 m2  b2 ) y=–x±5  x+y+5=0  minimum distance = distance between or x+ y– 5 =0 x + y – 10 = 0 & x + y – 5 = 0  shortest distance = |10  5 | = 11 Q P 5 2 ELLIPSE CHORD OF CONTACT If PA and PB be the tangents from point P(x1,y1) to the ellipse The equation of the chord of contact AB is Ex. x2 y2  2 1. 2 a b xx1 yy  21  1 or T = 0 (at x ,y ). 2 1 1 a b If tangents to the parabola y2 = 4ax intersect the ellipse x 2 y2   1 at A and B, the find the locus of point of a 2 b2 intersection of tangents at A and B. Sol. Let P  (h, k) be the point of intersection of tangents at A & B  equation of chord of contact AB is xh yk  1 a 2 b2 .......... (i) which touches the parabola. Equation of tangent to parabola y2 = 4ax is y = mx +  mx – y = – a m a m .......... (ii) equation (i) & (ii) as must be same  a  m 1   m 1  h   k   2  2 a b      – hb2 ak  2 ka 2 b P(h, k) h b2 ak ak & m  22 k aa22 bb  m=–  locus of P is y2 = –  b4 .x a3 EQUATION OF CHORD WITH MID POINT (x1,y1) The equation of the chord of the ellipse x2 y2  2  1 , whose mid-point be (x1,y1) is T = S1 2 a b x 12 y 12 x x 1 yy 1 S =  1, ,   1 1 2 a b2 a2 b2 where T = i.e. 2  y 12  xx 1 yy 1   x1  2 + 2  1 =  2  2  1 b b  a  a  Ex. Find the locus of the mid-point of focal chords of the ellipse Sol. Let P  (h, k) be the mid-point  equation of chord whose mid-point is given x 2 y2  1 . a 2 b2 xh yk h2 k2   1   1 a 2 b2 a 2 b2 since it is a focal chord,  It passes through focus, either (ae, 0) or (–ae, 0) If it passes through (ae, 0)  locus is ex x2 y2  2  2 a a b S If it passes through (–ae, 0)  locus is – P(h, k) ex x2 y2  2  2 a a b DIRECTOR CIRCLE Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle. The equation to this locus is x² + y² = a² + b² i.e. a circle whose centre is the centre of the ellipse & whose radius is the length of the line joining the ends of the major & minor axes. Ex. An ellipse slides between two perpendicular lines. Show that the locus of its centre is a circle. Sol. Let length of semi-major and semi-minor axis are 'a' and 'b' and centre is C  (h, k) Since ellipse slides between two perpendicular lines, therefore point of intersection of two perpendicular tangents lies on director circle. Let us consider two perpendicular lines as x & y axes  point of intersection is origin O  (0, 0)  OC = radius of director circle  h2  k2 = a 2  b2  locus of C  (h, k) is  x2 + y2 = a2 + b2 which is a circle C O IMPORTANT CONCEPTS (1) x 2 y2 =1  a 2 b2 If P be any point on the ellipse with S & S as its foci then (2) The tangent & normal at a point P on the ellipse bisect the external & internal angles between the focal Refering to the ellipse  (SP) +  (SP) = 2a. distances of P. This refers to the well known reflection property of the ellipse which states that rays from one focus are reflected through other focus & viceversa. Hence we can deduce that the straight lines joining each focus to the foot of the perpendicular from the other focus upon the tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis. ELLIPSE (3) The product of the length’s of the perpendicular segments from the foci on any tangent to the ellipse is b² and the feet of these perpendiculars lie on its auxiliary circle and the tangents at these feet to the auxiliary circle meet on the ordinate of P and that the locus of their point of intersection is a similiar ellipse as that of the original one. (4) The portion of the tangent to an ellipse between the point of contact & the directrix subtends a right angle at the corresponding focus. (5) If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively, & if CF be perpendicular upon this normal, then (i) PF. PG = b² (ii) PF. Pg = a² (iii) PG. Pg = SP. S P (iv) CG. CT = CS2 (v) locus of the mid point of Gg is another ellipse having P C G S F S g the same eccentricity as that of the original ellipse. [where S and S are the focii of the ellipse and T is the point where tangent at P meet the major axis] (6) The circle on any focal distance as diameter touches the auxiliary circle. Perpendiculars from the centre upon all chords which join the ends of any perpendicular diameters of the ellipse are of constant length. (7) If the tangent at the point P of a standard ellipse meets the axis in T and t and CY is the perpendicular on it from the centre then, (i) T t. PY = a2  b 2 and (ii) (8) least value of T t is a + b. Atmost four normals & two tangents can be drawn from any point to an ellipse. 1. Standard Equation & Definition Standard equation of an ellipse referred to its principal axis along the coordinate axis is where a > b & b² = a² (1  e²).  x 2 y2 = 1,  a 2 b2 a 2 – b 2 = a 2 e 2. where e = Eccentricity (0 < e < 1) FOCI : S  (a e, 0) & S ( a e, 0). a a &x= . e e (a) Equations of Directrices : x = (b) Vertices : A ( a, 0) & A  (a, 0) . (c) Major Axis : The line segment AA in which the foci S & S lie is of length 2a & is called the major axis (a > b) of the ellipse. Point of intersection of major axis with directrix is called  a  the foot of the directrix (Z)  ± , 0  .  e  (d) Minor Axis : The yaxis intersects the ellipse in the points B  (0,  b) & B  (0, b). The line segment BB is of length 2b (b < a) is called the Minor Axis of the ellipse. (e) Principal Axis : The major & minor axis together are called Principal Axis of the ellipse. (f) Centre : The point which bisects every chord of the conic drawn through it, is called the centre of the conic. C  (0, 0) the origin is the centre of the ellipse x y a b 2 2 2 2 = 1. (g) Diameter : A chord of the conic which passes through the centre is called a diameter of the conic. (h) Focal Chord : A chord which passes through a focus is called a focal chord. (i) Double Ordinate : A chord perpendicular to the major axis is called a double ordinate with respect to major axis as diameter. (j) Latus Rectum : The focal chord perpendicular to the major axis is called the latus rectum.  minor axis   2a 1  e2 2b2    Length of latus rectum (LL) = a major axis Equation of latus rectum : x =  ae. 2 (i) (ii) (iii)    b2  b2  b2   b2  Ends of the latus rectum are L  ae,  , L'  ae, –  , L1  –ae,  and L1'  –ae, –  a  a  a  a      ELLIPSE (k) Focal radii : SP = a – ex and S’P = a + ex  (l) 2. SP + S’P = 2a = Major axis Eccentricity : e = 1  b2 a2 Another form of Ellipse x  y = 1, (a < b) a b 2 2 2 2 (a) AA’ = Minor axis = 2a (b) BB’ = Major axis = 2b (c) a2 = b2 (1 – e2) (d) latus rectum LL' = L1L1' = 2a 2 . b equation y = ± be (e) Ends of the latus rectum are :  a2   a2   a2   a2  L  , be  , L'  – , be  , L1  , –be  , L1'  – , –be  b b b b         (f) 3. b e Equation of directrix y = ± . (g) Eccentricity : e = 1 – a2 b2 General Equation of an Ellipse Let (a,b) be the focus S, and lx + my + n = 0 is the equation of directrix. Let P (x, y) be any point on the ellipse. Then by definition.  SP = e PM (e is the eccentricity) (x – a) 2 + (y – b) 2 = e2  (l2 + m2) {(x-a)2 + (y-b)2} = e2{lx + my + n)2 M P(x,y) axis Position of a Point W.R.T. an Ellipse S(a,b) The point P(x1, y1) lies outside, inside or on the ellipse according as : x12 y12 + – 1 >< or = 0 a 2 b2 Y 5. Auxilliary Circle/eccentric Angle Q A circle described on major axis as diameter is called the auxilliary P circle. Let Q be a point on the auxilliary circle x2 + y2 = a2 such that QP produced is perpendicular to the x-axis then P & Q are called as the CORRESPONDING POINTS on the ellipse & the auxilliary circle respectively. ‘q’ is called ECCENTRIC ANGLE of the point P on the ellipse (0  0 < 2 ). A’ (-a,0)  S’ O N S A (a,0) lx + my + n = 0 4. (lx + my + n) 2 (l2 + m 2 )  Note that l(PN) b Semi minor axis = = l(QN) a Semi major axis Hence “If from each point of a circle perpendiculars are drawn upon a fixed diameter then the locus of the points dividing these perpendiculars in a given ratio is an ellipse of which the given circle is the auxilliary circle.” 6. Parametric Representation The equations x = a cos  & y = b sin  together represent the ellipse x 2 y2 + =1 a 2 b2 where  is a parameter (eccentric angle). Note that if P() = (a cos , b sin ) is on the ellipse then ; Q() = (a cos , a sin ) is on the auxilliary circle. 7. Line and an Ellipse The line y = mx + c meets the ellipse x 2 y2 + = 1 in two real points, coincident or imaginary according as c2 is a 2 b2 < = or > a2m2 + b 2. Hence y = mx + c is tangent to the ellipse x 2 y2 + = 1 if c2 = a2m2 + b2. a 2 b2 The equation to the chord of the ellipse joining two points with eccentric angles  &  is given by x α +β y α +β α -β cos + sin = cos . a 2 b 2 2 8. Tangent to the Ellipse (a) x 2 y2 + =1 : a 2 b2 Point form : Equation of tangent to the given ellipse at its point (x1, y1) is xx1 yy1 + 2 =1 a2 b (b) Slope form : Equation of tangent to the given ellipse whose slope is ‘m’, is y = mx ± a 2 m 2 + b 2  ±a 2 m mb2 ,  Point of contact are  2 2 2 a 2 m 2 + b2  a m +b (c)    Parametric form : Equation of tangent to given ellipse at its point (a cos , b sin ), is xcosθ ysinθ + =1 a b ELLIPSE 9. Normal to the Ellipse (a) Point form : Equation of the normal to the given ellipse at (x1, y1) is (b) a 2 x b2 y – = a 2 – b 2 = a 2e2 . x1 y Slope form : Equation of a normal to the given ellipse whose slope is ‘m’ is y = mx + (c) x 2 y2 + =1 a 2 b2 (a 2 – b 2 )m a 2 + b2m2 Parametric form : Equation of the normal to the given ellipse at the point (a cos, b sin) is ax . sec  – by . cosec  = (a2 – b2). 10. Chord of Contact If PA and PB be the tangents from point P(x1,y1) to the ellipse x 2 y2 xx yy + = 1 , then the equation of the chord of contact AB is 21 + 21 = 1 or T = 0 at (x1.y1) a 2 b2 a b 11. Pair of Tangents Y If P(x1, y1) be any points lies outside the ellipse x 2 y2 + = 1 , and a a 2 b2 pair of tangents PA, PB can be drawn to it from P. Then the equation of pair of tangents of PA and PB is SS1 = T 2 2 where S1 = x1 a 2 + A P (x1,y1) X’ C X B xx1 yy1 y12 –1 , T = 2 + 2 – 1 a b b2 Y’ 2 2  x 2 y 2   x 1 y12   xx1 yy1  i.e.  2 + 2 –1  2 + 2 –1 =  2 + 2 –1  b b b  a  a   a 12. Director Circle Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle. The equation to this locus is x2 + y2 = a2 + b2 i.e. a circle whose centre is the centre of the ellipse & whose radius is the length of the line joining the ends of the major & minor axis. 13. Equation of Chord with Mid Point (x1, y1) The equation of the chord of the ellipse where T = x 21 y12 xx1 yy1 , + –1 S = + –1 1 a2 b2 a 2 b2 2 2   xx1 yy1   x 1 y1 + –1 = i.e.  2   2 + 2 –1  2 b b  a  a  x 2 y2 + = 1 , whose mid-point be (x1, y1) is T = S1 a 2 b2