SlideShare una empresa de Scribd logo
1 de 29
Descargar para leer sin conexión
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

Más contenido relacionado

Similar a 11X1 T12 08 geometrical theorems (2011)

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
 

Similar a 11X1 T12 08 geometrical theorems (2011) (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
 

Más 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
 

Más 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

The basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptxThe basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptx
heathfieldcps1
 
ppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyesppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyes
ashishpaul799
 

Último (20)

“O BEIJO” EM ARTE .
“O BEIJO” EM ARTE                       .“O BEIJO” EM ARTE                       .
“O BEIJO” EM ARTE .
 
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdfPost Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
 
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdfTelling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
 
The Last Leaf, a short story by O. Henry
The Last Leaf, a short story by O. HenryThe Last Leaf, a short story by O. Henry
The Last Leaf, a short story by O. Henry
 
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General QuizPragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
Pragya Champions Chalice 2024 Prelims & Finals Q/A set, General Quiz
 
The basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptxThe basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptx
 
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
 
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdfDanh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
 
Matatag-Curriculum and the 21st Century Skills Presentation.pptx
Matatag-Curriculum and the 21st Century Skills Presentation.pptxMatatag-Curriculum and the 21st Century Skills Presentation.pptx
Matatag-Curriculum and the 21st Century Skills Presentation.pptx
 
slides CapTechTalks Webinar May 2024 Alexander Perry.pptx
slides CapTechTalks Webinar May 2024 Alexander Perry.pptxslides CapTechTalks Webinar May 2024 Alexander Perry.pptx
slides CapTechTalks Webinar May 2024 Alexander Perry.pptx
 
factors influencing drug absorption-final-2.pptx
factors influencing drug absorption-final-2.pptxfactors influencing drug absorption-final-2.pptx
factors influencing drug absorption-final-2.pptx
 
How to Manage Notification Preferences in the Odoo 17
How to Manage Notification Preferences in the Odoo 17How to Manage Notification Preferences in the Odoo 17
How to Manage Notification Preferences in the Odoo 17
 
ppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyesppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyes
 
....................Muslim-Law notes.pdf
....................Muslim-Law notes.pdf....................Muslim-Law notes.pdf
....................Muslim-Law notes.pdf
 
Behavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdfBehavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdf
 
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/
 
Essential Safety precautions during monsoon season
Essential Safety precautions during monsoon seasonEssential Safety precautions during monsoon season
Essential Safety precautions during monsoon season
 
Basic Civil Engg Notes_Chapter-6_Environment Pollution & Engineering
Basic Civil Engg Notes_Chapter-6_Environment Pollution & EngineeringBasic Civil Engg Notes_Chapter-6_Environment Pollution & Engineering
Basic Civil Engg Notes_Chapter-6_Environment Pollution & Engineering
 
[GDSC YCCE] Build with AI Online Presentation
[GDSC YCCE] Build with AI Online Presentation[GDSC YCCE] Build with AI Online Presentation
[GDSC YCCE] Build with AI Online Presentation
 
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
UNIT – IV_PCI Complaints: Complaints and evaluation of complaints, Handling o...
 

11X1 T12 08 geometrical theorems (2011)

  • 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