SlideShare uma empresa Scribd logo
1 de 29
Baixar para ler offline
Geometrical Theorems about
         Parabola
Geometrical Theorems about
(1) Focal Chords
                 Parabola
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
                Tangents are perpendicular to each other
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
                  Tangents are perpendicular to each other
 3 Show that the point of intersection,T , of the tangents is
     a  p  q  , apq
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
                Tangents are perpendicular to each other
 3 Show that the point of intersection,T , of the tangents is
   a  p  q  , apq                 y  apq
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
                Tangents are perpendicular to each other
 3 Show that the point of intersection,T , of the tangents is
   a  p  q  , apq                 y  apq
                                      y  a       pq  1
Geometrical Theorems about
 (1) Focal Chords
                  Parabola
e.g. Prove that the tangents drawn from the extremities of a focal chord
     intersect at right angles on the directrix.
 1 Prove pq  1

 2 Show that the slope of the tangent at P is p, and the slope of the
   tangent at Q is q.
                             pq  1
                Tangents are perpendicular to each other
 3 Show that the point of intersection,T , of the tangents is
   a  p  q  , apq                 y  apq
                                      y  a      pq  1
                                Tangents meet on the directrix
(2) Reflection Property
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
                                         Data: CP || y axis
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
                                         Data: CP || y axis

                                    1 Show tangent at P is y  px  ap 2
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
                                         Data: CP || y axis

                                    1 Show tangent at P is y  px  ap 2

                                    2 tangent meets y axis when x = 0
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
                                         Data: CP || y axis

                                    1 Show tangent at P is y  px  ap 2

                                    2 tangent meets y axis when x = 0
                                              K is 0,ap 2 
(2) Reflection Property
Any line parallel to the axis of the parabola is reflected towards the
focus.
Any line from the focus parallel to the axis of the parabola is reflected
parallel to the axis.
Thus a line and its reflection are equally inclined to the normal, as well
as to the tangent.
                                         Prove: SPK  CPB
                                (angle of incidence = angle of reflection)
                                         Data: CP || y axis

                                    1 Show tangent at P is y  px  ap 2

                                    2 tangent meets y axis when x = 0
                                              K is 0,ap 2 
                                               d SK  a  ap 2
2ap  0  ap  a 
                             2
d PS 
                 2      2
2ap  0  ap  a 
                                  2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1
2ap  0  ap  a 
                                  2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
2ap  0  ap  a 
                                  2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
    SPK is isosceles             two = sides 
2ap  0  ap  a 
                                  2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
    SPK is isosceles             two = sides 
     SPK  SKP (base 's isosceles  )
2ap  0  ap  a 
                                      2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
    SPK is isosceles                 two = sides 
     SPK  SKP (base 's isosceles  )
     SKP  CPB                  (corresponding 's  , SK || CP)
2ap  0  ap  a 
                                      2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
    SPK is isosceles                 two = sides 
     SPK  SKP (base 's isosceles  )
     SKP  CPB                  (corresponding 's  , SK || CP)
    SPK  CPB
2ap  0  ap  a 
                                      2
d PS 
                      2       2


     4a 2 p 2  a 2 p 4  2a 2 p 2  a 2
     a p4  2 p2  1

          p        1
                          2
    a         2



     a  p 2  1  d SK
    SPK is isosceles                 two = sides 
     SPK  SKP (base 's isosceles  )
     SKP  CPB                  (corresponding 's  , SK || CP)
    SPK  CPB


                    Exercise 9I; 1, 2, 4, 7, 11, 12, 17, 18, 21

Mais conteúdo relacionado

Destaque

11X1 T16 03 polynomial division
11X1 T16 03 polynomial division11X1 T16 03 polynomial division
11X1 T16 03 polynomial division
Nigel Simmons
 
11X1 T06 04 probability and counting techniques (2010)
11X1 T06 04 probability and counting techniques (2010)11X1 T06 04 probability and counting techniques (2010)
11X1 T06 04 probability and counting techniques (2010)
Nigel Simmons
 
11X1 T15 01 applications of ap & gp
11X1 T15 01 applications of ap & gp11X1 T15 01 applications of ap & gp
11X1 T15 01 applications of ap & gp
Nigel Simmons
 
X1 T4 2 angles of any magnitude (2010)
X1 T4 2 angles of any magnitude (2010)X1 T4 2 angles of any magnitude (2010)
X1 T4 2 angles of any magnitude (2010)
Nigel Simmons
 
X2 T05 01 by parts (2010)
X2 T05 01 by parts (2010)X2 T05 01 by parts (2010)
X2 T05 01 by parts (2010)
Nigel Simmons
 
X2 T01 09 geometrical representation (2010)
X2 T01 09 geometrical representation (2010)X2 T01 09 geometrical representation (2010)
X2 T01 09 geometrical representation (2010)
Nigel Simmons
 
X2 T07 02 resisted motion (2010)
X2 T07 02 resisted motion (2010)X2 T07 02 resisted motion (2010)
X2 T07 02 resisted motion (2010)
Nigel Simmons
 
11X1 T05 04 point slope formula (2010)
11X1 T05 04 point slope formula (2010)11X1 T05 04 point slope formula (2010)
11X1 T05 04 point slope formula (2010)
Nigel Simmons
 
11X1 T13 01 definitions & chord theorems (2010)
11X1 T13 01 definitions & chord theorems (2010)11X1 T13 01 definitions & chord theorems (2010)
11X1 T13 01 definitions & chord theorems (2010)
Nigel Simmons
 

Destaque (9)

11X1 T16 03 polynomial division
11X1 T16 03 polynomial division11X1 T16 03 polynomial division
11X1 T16 03 polynomial division
 
11X1 T06 04 probability and counting techniques (2010)
11X1 T06 04 probability and counting techniques (2010)11X1 T06 04 probability and counting techniques (2010)
11X1 T06 04 probability and counting techniques (2010)
 
11X1 T15 01 applications of ap & gp
11X1 T15 01 applications of ap & gp11X1 T15 01 applications of ap & gp
11X1 T15 01 applications of ap & gp
 
X1 T4 2 angles of any magnitude (2010)
X1 T4 2 angles of any magnitude (2010)X1 T4 2 angles of any magnitude (2010)
X1 T4 2 angles of any magnitude (2010)
 
X2 T05 01 by parts (2010)
X2 T05 01 by parts (2010)X2 T05 01 by parts (2010)
X2 T05 01 by parts (2010)
 
X2 T01 09 geometrical representation (2010)
X2 T01 09 geometrical representation (2010)X2 T01 09 geometrical representation (2010)
X2 T01 09 geometrical representation (2010)
 
X2 T07 02 resisted motion (2010)
X2 T07 02 resisted motion (2010)X2 T07 02 resisted motion (2010)
X2 T07 02 resisted motion (2010)
 
11X1 T05 04 point slope formula (2010)
11X1 T05 04 point slope formula (2010)11X1 T05 04 point slope formula (2010)
11X1 T05 04 point slope formula (2010)
 
11X1 T13 01 definitions & chord theorems (2010)
11X1 T13 01 definitions & chord theorems (2010)11X1 T13 01 definitions & chord theorems (2010)
11X1 T13 01 definitions & chord theorems (2010)
 

Semelhante a 11X1 T11 08 geometrical theorems (2010)

11 x1 t11 08 geometrical theorems
11 x1 t11 08 geometrical theorems11 x1 t11 08 geometrical theorems
11 x1 t11 08 geometrical theorems
Nigel Simmons
 
11 x1 t11 09 locus problems (2013)
11 x1 t11 09 locus problems (2013)11 x1 t11 09 locus problems (2013)
11 x1 t11 09 locus problems (2013)
Nigel Simmons
 
The shortest distance between skew lines
The shortest distance between skew linesThe shortest distance between skew lines
The shortest distance between skew lines
Tarun Gehlot
 
Inmo 2013 test_paper_solution
Inmo 2013 test_paper_solutionInmo 2013 test_paper_solution
Inmo 2013 test_paper_solution
Suresh Kumar
 
Thermodynamics of crystalline states
Thermodynamics of crystalline statesThermodynamics of crystalline states
Thermodynamics of crystalline states
Springer
 
Thermodynamics of crystalline states
Thermodynamics of crystalline statesThermodynamics of crystalline states
Thermodynamics of crystalline states
Springer
 
Analisis Korespondensi
Analisis KorespondensiAnalisis Korespondensi
Analisis Korespondensi
dessybudiyanti
 
Cylindrical and spherical coordinates
Cylindrical and spherical coordinatesCylindrical and spherical coordinates
Cylindrical and spherical coordinates
Tarun Gehlot
 
Cylindrical and spherical coordinates
Cylindrical and spherical coordinatesCylindrical and spherical coordinates
Cylindrical and spherical coordinates
Tarun Gehlot
 
Parabola
ParabolaParabola
Parabola
itutor
 

Semelhante a 11X1 T11 08 geometrical theorems (2010) (20)

11 x1 t11 08 geometrical theorems
11 x1 t11 08 geometrical theorems11 x1 t11 08 geometrical theorems
11 x1 t11 08 geometrical theorems
 
11 x1 t11 09 locus problems (2013)
11 x1 t11 09 locus problems (2013)11 x1 t11 09 locus problems (2013)
11 x1 t11 09 locus problems (2013)
 
Two_variations_on_the_periscope_theorem.pdf
Two_variations_on_the_periscope_theorem.pdfTwo_variations_on_the_periscope_theorem.pdf
Two_variations_on_the_periscope_theorem.pdf
 
Curve generation %a1 v involute and evolute
Curve generation %a1 v involute and evoluteCurve generation %a1 v involute and evolute
Curve generation %a1 v involute and evolute
 
F Giordano Collins Fragmentation for Kaon
F Giordano Collins Fragmentation for KaonF Giordano Collins Fragmentation for Kaon
F Giordano Collins Fragmentation for Kaon
 
COORDINATE GEOMETRY II
COORDINATE GEOMETRY IICOORDINATE GEOMETRY II
COORDINATE GEOMETRY II
 
The shortest distance between skew lines
The shortest distance between skew linesThe shortest distance between skew lines
The shortest distance between skew lines
 
class 10 circles
class 10 circlesclass 10 circles
class 10 circles
 
Inmo 2013 test_paper_solution
Inmo 2013 test_paper_solutionInmo 2013 test_paper_solution
Inmo 2013 test_paper_solution
 
Thermodynamics of crystalline states
Thermodynamics of crystalline statesThermodynamics of crystalline states
Thermodynamics of crystalline states
 
Thermodynamics of crystalline states
Thermodynamics of crystalline statesThermodynamics of crystalline states
Thermodynamics of crystalline states
 
Analisis Korespondensi
Analisis KorespondensiAnalisis Korespondensi
Analisis Korespondensi
 
Cylindrical and spherical coordinates
Cylindrical and spherical coordinatesCylindrical and spherical coordinates
Cylindrical and spherical coordinates
 
Cylindrical and spherical coordinates
Cylindrical and spherical coordinatesCylindrical and spherical coordinates
Cylindrical and spherical coordinates
 
Curves part two
Curves part twoCurves part two
Curves part two
 
Parabola
ParabolaParabola
Parabola
 
Permuting Polygons
Permuting PolygonsPermuting Polygons
Permuting Polygons
 
Planar projective geometry
Planar projective geometryPlanar projective geometry
Planar projective geometry
 
Hybridization
HybridizationHybridization
Hybridization
 
Parabola
ParabolaParabola
Parabola
 

Mais de Nigel Simmons

12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
Nigel Simmons
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
Nigel Simmons
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
Nigel Simmons
 

Mais de Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 

Último

MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
Krashi Coaching
 
Financial Accounting IFRS, 3rd Edition-dikompresi.pdf
Financial Accounting IFRS, 3rd Edition-dikompresi.pdfFinancial Accounting IFRS, 3rd Edition-dikompresi.pdf
Financial Accounting IFRS, 3rd Edition-dikompresi.pdf
MinawBelay
 

Último (20)

ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
 
“O BEIJO” EM ARTE .
“O BEIJO” EM ARTE                       .“O BEIJO” EM ARTE                       .
“O BEIJO” EM ARTE .
 
REPRODUCTIVE TOXICITY STUDIE OF MALE AND FEMALEpptx
REPRODUCTIVE TOXICITY  STUDIE OF MALE AND FEMALEpptxREPRODUCTIVE TOXICITY  STUDIE OF MALE AND FEMALEpptx
REPRODUCTIVE TOXICITY STUDIE OF MALE AND FEMALEpptx
 
How to Analyse Profit of a Sales Order in Odoo 17
How to Analyse Profit of a Sales Order in Odoo 17How to Analyse Profit of a Sales Order in Odoo 17
How to Analyse Profit of a Sales Order in Odoo 17
 
IPL Online Quiz by Pragya; Question Set.
IPL Online Quiz by Pragya; Question Set.IPL Online Quiz by Pragya; Question Set.
IPL Online Quiz by Pragya; Question Set.
 
An Overview of the Odoo 17 Knowledge App
An Overview of the Odoo 17 Knowledge AppAn Overview of the Odoo 17 Knowledge App
An Overview of the Odoo 17 Knowledge App
 
ANTI PARKISON DRUGS.pptx
ANTI         PARKISON          DRUGS.pptxANTI         PARKISON          DRUGS.pptx
ANTI PARKISON DRUGS.pptx
 
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
 
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
 
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
 
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
 Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
 
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
 
MOOD STABLIZERS DRUGS.pptx
MOOD     STABLIZERS           DRUGS.pptxMOOD     STABLIZERS           DRUGS.pptx
MOOD STABLIZERS DRUGS.pptx
 
HVAC System | Audit of HVAC System | Audit and regulatory Comploance.pptx
HVAC System | Audit of HVAC System | Audit and regulatory Comploance.pptxHVAC System | Audit of HVAC System | Audit and regulatory Comploance.pptx
HVAC System | Audit of HVAC System | Audit and regulatory Comploance.pptx
 
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
 
An overview of the various scriptures in Hinduism
An overview of the various scriptures in HinduismAn overview of the various scriptures in Hinduism
An overview of the various scriptures in Hinduism
 
Championnat de France de Tennis de table/
Championnat de France de Tennis de table/Championnat de France de Tennis de table/
Championnat de France de Tennis de table/
 
II BIOSENSOR PRINCIPLE APPLICATIONS AND WORKING II
II BIOSENSOR PRINCIPLE APPLICATIONS AND WORKING IIII BIOSENSOR PRINCIPLE APPLICATIONS AND WORKING II
II BIOSENSOR PRINCIPLE APPLICATIONS AND WORKING II
 
Dementia (Alzheimer & vasular dementia).
Dementia (Alzheimer & vasular dementia).Dementia (Alzheimer & vasular dementia).
Dementia (Alzheimer & vasular dementia).
 
Financial Accounting IFRS, 3rd Edition-dikompresi.pdf
Financial Accounting IFRS, 3rd Edition-dikompresi.pdfFinancial Accounting IFRS, 3rd Edition-dikompresi.pdf
Financial Accounting IFRS, 3rd Edition-dikompresi.pdf
 

11X1 T11 08 geometrical theorems (2010)

  • 2. Geometrical Theorems about (1) Focal Chords Parabola
  • 3. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix.
  • 4. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1
  • 5. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q.
  • 6. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1
  • 7. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1 Tangents are perpendicular to each other
  • 8. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1 Tangents are perpendicular to each other 3 Show that the point of intersection,T , of the tangents is a  p  q  , apq
  • 9. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1 Tangents are perpendicular to each other 3 Show that the point of intersection,T , of the tangents is a  p  q  , apq y  apq
  • 10. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1 Tangents are perpendicular to each other 3 Show that the point of intersection,T , of the tangents is a  p  q  , apq y  apq  y  a  pq  1
  • 11. Geometrical Theorems about (1) Focal Chords Parabola e.g. Prove that the tangents drawn from the extremities of a focal chord intersect at right angles on the directrix. 1 Prove pq  1 2 Show that the slope of the tangent at P is p, and the slope of the tangent at Q is q. pq  1 Tangents are perpendicular to each other 3 Show that the point of intersection,T , of the tangents is a  p  q  , apq y  apq  y  a  pq  1 Tangents meet on the directrix
  • 13. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent.
  • 14. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent.
  • 15. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB
  • 16. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection)
  • 17. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection) Data: CP || y axis
  • 18. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection) Data: CP || y axis 1 Show tangent at P is y  px  ap 2
  • 19. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection) Data: CP || y axis 1 Show tangent at P is y  px  ap 2 2 tangent meets y axis when x = 0
  • 20. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection) Data: CP || y axis 1 Show tangent at P is y  px  ap 2 2 tangent meets y axis when x = 0  K is 0,ap 2 
  • 21. (2) Reflection Property Any line parallel to the axis of the parabola is reflected towards the focus. Any line from the focus parallel to the axis of the parabola is reflected parallel to the axis. Thus a line and its reflection are equally inclined to the normal, as well as to the tangent. Prove: SPK  CPB (angle of incidence = angle of reflection) Data: CP || y axis 1 Show tangent at P is y  px  ap 2 2 tangent meets y axis when x = 0  K is 0,ap 2  d SK  a  ap 2
  • 22. 2ap  0  ap  a  2 d PS  2 2
  • 23. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1
  • 24. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK
  • 25. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK SPK is isosceles  two = sides 
  • 26. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK SPK is isosceles  two = sides  SPK  SKP (base 's isosceles  )
  • 27. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK SPK is isosceles  two = sides  SPK  SKP (base 's isosceles  ) SKP  CPB (corresponding 's  , SK || CP)
  • 28. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK SPK is isosceles  two = sides  SPK  SKP (base 's isosceles  ) SKP  CPB (corresponding 's  , SK || CP)  SPK  CPB
  • 29. 2ap  0  ap  a  2 d PS  2 2  4a 2 p 2  a 2 p 4  2a 2 p 2  a 2  a p4  2 p2  1 p  1 2 a 2  a  p 2  1  d SK SPK is isosceles  two = sides  SPK  SKP (base 's isosceles  ) SKP  CPB (corresponding 's  , SK || CP)  SPK  CPB Exercise 9I; 1, 2, 4, 7, 11, 12, 17, 18, 21