## Mimi_x3 4 years ago If the chord of contact of the tangents to a parabola x^2=4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus.

1. perl

mmmmm, you werent kidding around

2. Mimi_x3

Is this easy for you ? xD

3. perl

i will have to think about it

4. perl

the directrix is that line under the parabola

5. Mimi_x3

lols, you were thinking that its easy ? xD

6. Mimi_x3

Yup, y=-a

7. perl

what class is this ?

8. Mimi_x3

I live in Australia, so its kinda different, there's only a maths class xD

9. perl

sorry wrong question, have you done calculus

10. Mimi_x3

Yes I have, how did you know ? xD

11. Mimi_x3

What grade are you in ?

12. perl

it will definitely help

13. perl

im a sophomore in college

14. Mimi_x3

Cool~ college, and you havent learnt that ?

15. perl

not specifically, but i bet i can do this

16. Mimi_x3

lols, then try hehe

17. perl

im trying to make a diagram, im not even sure i understand the question yet

18. perl

ok you have an external pt, and you want the tangent through it

19. Mimi_x3

Yeah, I don't really know where's the external point is, thats why im lost =/

20. perl

ok first thing is first. the chord of contact means something special

21. Mimi_x3

Something special ? What do you mean ?

22. perl

there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

23. perl

well given an external point there are two tangents from that point to the parabola

24. perl

now where the tangents intersect the parabola, now draw a chord through those two points

25. perl

where the tangents meet the parabola, the two tangents

26. Mimi_x3

wow~ nice pdf file, its detailed on parametrics ty

27. Mimi_x3

But where is the external point ?

28. perl

anywhere

29. Mimi_x3

Can't it have to be specific

30. perl

in the pdf, the point is below the parabola

31. Mimi_x3

It can be but it doesn't make sense when it says RT subtends a right angle at the focus.

32. perl

like this

33. Mimi_x3

like this ?

34. perl

|dw:1319452686670:dw|

35. Mimi_x3

I know but where's the external point ?

36. perl

|dw:1319452750463:dw|

37. Mimi_x3

Imaginary ?

38. perl

|dw:1319452790390:dw|

39. Mimi_x3

Is the parabola imaginery or somethinbg ?

40. Mimi_x3

imaginary* something*

41. perl

|dw:1319452894222:dw|

42. Mimi_x3

I don't get your diagrams ? sorry

43. perl

|dw:1319453015665:dw|

44. Mimi_x3

I don't get it ? >_<

45. perl

yeah the diagram is a beast

46. Mimi_x3

lols, why is there like 3 digrams ? i dont get it >_<

47. perl

|dw:1319453240562:dw|

48. Mimi_x3

lols, this is not 3D trig xD

49. perl

can you do better ?

50. Mimi_x3

Nope, i dont know how to start, the problem is where is T

51. perl

just a point below the parabola

52. perl

like in that pdf

53. Mimi_x3

But it won't make sense, and i dont know how to prove

54. perl

have faith

55. Mimi_x3

have faith ?

56. perl

i think they mean the segment R T

57. perl

yes have faith when you draw it that it will all come into place, the problem will unravel

58. perl

i am atheist, what did you think i mean ?

59. Mimi_x3

idk, never mind then i will work it out myself

60. perl

|dw:1319453605908:dw|

61. Mimi_x3

ty (:

62. perl

im working on it

63. perl

would be nice if they gave you a diagram

64. Mimi_x3

Yeah i know, none of the hard questions have diagrams =/

65. perl

i think RT is the line segment though

66. Mimi_x3

hm idk

67. perl

|dw:1319454238070:dw|

68. anonymous

$\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}$ $x^2=4py$

69. anonymous

so the directrix would be y = -a in your question :-D

70. perl

yes

71. Mimi_x3

Yeah, but im having problem proving

72. anonymous

If the chord of contact of the tangents to a parabola x² = 4ay, from an external point T(x1, y1) meets the directrix at R, prove that RT subtends a right angle at the focus. chord of contact of tangents to parabola is xx1=2a(y+y1) directrix is y=-a substitute into that, x=[2a(-a+y1)]/x1 y=-a R([2a(-a+y1)]/x1,-a) let S be the focus d/dx RS = 2a/ -[2a(-a+y1)]/x1= x1/(a-y1) d/dx TS=(y1-a)/x1 d/dx RS * d/dx TS= -1

73. Mimi_x3

lols, you got it from boredofstudies right ?

74. anonymous

yes... :-P

75. perl

how did you get chord of tangents equation

76. perl

hey, can you draw a diagram then of whats going on, please post the link

77. Mimi_x3

Yeah but the problem is that I don't understand the solution

78. anonymous

it helps to draw a diagram.

79. perl

80. anonymous
81. Mimi_x3

But what's xx1 ? im trying to figure it out

82. anonymous
83. Mimi_x3

oh i get it, but where did y=1/2(p+q)x-apq come from ?

84. anonymous

xx1 must mean something like $x_1\times x= 2a(y+y_1)$

85. Mimi_x3

where did 2a come from ?

86. anonymous

good question

87. anonymous

$parabola: x^2 = 4py$$a = \frac{1}{4p}$

88. Mimi_x3

no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a

89. perl

|dw:1319455562515:dw|

90. perl

i wish there were other colors, but from T draw two tangents to the parabola

91. anonymous

perl's diagram works.... but how do we get there using calculus and derivatives if we don't even know the equation for chord of contact of the tangents to a parabola

92. Mimi_x3

oh the equation of the tangent to the parabola is y=px-ap^2 right ?

93. perl

|dw:1319455702848:dw|

94. perl

they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

95. perl

the chords of tangents formula

96. anonymous

now it all makes sense

97. perl

|dw:1319455885258:dw|

98. perl

i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p

99. Mimi_x3

Doesn't it involve parameters as well P(2ap,ap^2)

100. anonymous

they want something like the quotient of the slopes between 2 lines is equal to -1

101. anonymous

and one of those lines should be RT

102. perl

|dw:1319456032235:dw|

103. perl

crap, it missed the R it is the right bottom corner

104. Mimi_x3

Im confused, is parameters involved ?

105. anonymous

no parameter seems to be involved :-(

106. perl

actually a is the parameter, you have to be careful, this word parameter is used in different senses

107. perl

, for instance in y = mx + b , m and b are parameters . now another sense of parameter is used when we 'parametrize' an equation. this means we introduce a third variable, often t

108. Mimi_x3

isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?

109. anonymous

meh, I assumed m and b to be constant

110. perl

right, but they are considered 'parameters'

111. Mimi_x3

but its not a straight line so it cant be y=mx + b

112. perl

in the sense that they change your line . so for x^2 = 4ay, the parameter is a since that produces different parabolas by tweaking a

113. perl

think of in this sense a parameter is your tweaking constant

114. anonymous

:-(

115. perl

the parameter is 'a'*

116. anonymous

I thought a was just a slope

117. Mimi_x3

Im lost =/ Parameter P(2ap,ap^2)_

118. perl

changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line

119. perl

these are your tweaking values .

120. perl

I dont like their use of p, scratch that and use t instead

121. perl

ok they are using 'parametric equations' , x = f(t), y = g(t)

122. perl

so x = 2at, y = at^2

123. anonymous

I like to use matrices and vectors.

124. Mimi_x3

Yeah thats in cartesian form

125. perl

ok, use a matrix or vector, id like to see that

126. perl

mimi, when you remove t, you get cartesian equation

127. Mimi_x3

matrix or vector ?

128. Mimi_x3

Oh yeah my bad i forgot xD

129. anonymous
130. Mimi_x3

I haven't learnt that complicated stuff yet =/

131. anonymous

you will learn it soon.

132. Mimi_x3

Hope not looks hard already =/

133. perl

thats curve fitting, no big deal

134. anonymous

you will find it easier than conics

135. Mimi_x3

Whats conics ?

136. perl

its a quick way to fit a quadratic sequence, cubic, etc

137. perl

mimi, what we are doing is conics, oh dea

138. anonymous

conic sections: parabolas, circles, ellipses, hyperbolas, stuff.

139. perl

oh dear *

140. anonymous

you do them along parametric equations usually.

141. perl

agdg, put the proper definition on board ;)

142. anonymous

well, they're what you get when you cut a cone.

143. Mimi_x3

LOL, this is not conics

144. perl

a conic has a specific algebraic equation in x and y , quadratic Ax^2 + Bxy + cY^2 + dy + ex + f = 0

145. perl

yes it is, a parabola is a conic

146. perl

a conic is a family of curves, as agdg pointed out

147. Mimi_x3

But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?

148. anonymous

forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically

149. anonymous
150. perl

right

151. Mimi_x3

Um, this doesnt look like a conic

152. perl

agdg, whats the nth term of this quadratic sequence 3,5,10,18...

153. perl

154. anonymous

hmmm... is it infinity?

155. perl

no, the nth term

156. perl

give me a formula for the sequence

157. perl

obviously it diverges, ;)

158. anonymous

its $2^n + 2^\frac{n}{2}$

159. perl

no, lol

160. anonymous

wait! it's $\frac{2^n+2^{\frac{n}{2}}}{2}$

161. perl

162. anonymous

how do we find it :-D

163. anonymous

so what's the 5th term

164. perl

ok so first we find the differences

165. perl

the first row of differences is 2 , 5, 8, 11 , ... the second row of differences is 3, 3, 3, ... so you know its a quadratic sequence

166. anonymous

that looks like a long matrix

167. perl

3, 5, 10, 18, 29 ... thats the sequence

168. perl

it should fit f(n) = an^2 + bn + c

169. anonymous

I guess that works.

170. anonymous

how do we multiply matrices?

171. perl

are you asking in general ?