# P5 Hyperbola QuestionWatch

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#1
Can someone please show me how to do this question (its page 98, question 5 of the heinemann book)

P is any point on the hyperbola with equation x^2/a^2 - y^2/b^2 = 1, S is the focus (ae, 0) and S' is the focus (-ae, 0). Show that:

|SP - S'P| = 2a

0
14 years ago
#2
SP/PH = e, S'P/PH' = e
where H and H' are projections of P onto 2 diretrices.
SP - SP' = e(PH - PH').
Where PH = xP + a^2/c, PH' = xP - a^2/c
e|PH - PH'| = 2a^2e/c
We know e = c/a , so e|PH = PH'| = 2a^2.c/a.c = 2a
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#3
Thanks for your help. Unfortunately I am now stuck on another question:

Prove that the triangle formed by the asymptotes of the curve with equation x^2 - 2y^2 = 4 and any tangent to the curve is of constant area.
0
14 years ago
#4
Edit: I used the wrong eqn for the hyperbola here. It should be x² - 2y² = 4 instead of x² - y² = 4.
Sorry about that , but the method should still be ok.

I can't draw any diagrams yet, so I'll try to describe the diagram I've been working with.
The hyperbola is x² - y² = 4
The asymptotes are y=x and y = -x (see later)
I have drawn a tangent to the curve in the positive quadrant of the graph.
The tangent cuts the curve at P
The tangent cuts the line y=x at the point A.
The tangent cuts the line y = -x at the point B
The origin =is the point O.

For a hyperbola x²/a² - y²/b² = 1, the asymptotes are the lines y = ± (b/a)x

Since here, a = b = 1, then
the asymptotes are
y = x , and y = -x
=============

Our curve is,

x² - y² = 4

differentiating,

2x - 2y.dy/dx = 0
dy/dx = x/y
========

At the point P(x', y')

dy/dx = x'/y', i.e.
m = x'/y'
======

Eqn of the tangent

y - y' = m(x - x')
y = xx'/y' - x'²/y' + y'
yy' = xx' - x'² + y'²
yy' = xx' - x'² + y'²
=============

Intersection of tangent with the line y = x at the point A.
A
==
y = x
yy' = xx' - x'² + y'²

substituting for y=x,
xy' = xx' - x'² + y'²
x(x' - y') = x'² - y'²
x = x' + y', -> y = x' + y'
A is the point A(x' + y', x' + y')
=====================

Intersection of tangent with the line y = -x at the point B.
B
==
y = -x
yy' = xx' - x'² + y'²

substituting for y=-x,
-xy' = xx' - x'² + y'²
x(x' + y') = x'² - y'²
x = x' - y', -> y = y' + x'
B is the point A(x' - y', y' - x')
=====================

Area of triangle OAB is Ar = ½.OA.OB.sin(AOB)
AOB=90 deg
=> Ar = ½.OA.OB
==============

OA = √[(x' + y')² + (x' + y')²] = (x' + y')√2
OB = √[(x' - y')² + (y' - x')²] = (x' - y')√2

Ar = ½.(x' + y')√2.(x' - y')√2
Ar = (x' + y')(x' - y')
Ar = x'² - y²
Ar = 4 = const (since x'² - y² = 4)
===========
0
14 years ago
#5
Oh... small mistake from the beginning.
The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok
0
14 years ago
#6
(Original post by BCHL85)
Oh... small mistake from the beginning.
The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok
Aaargh, My eyes are dim

Sorry 'bout that. I should wear my specs more often,
Oh well, at least the method is there
0
14 years ago
#7
(Original post by Fermat)
Aaargh, My eyes are dim

Sorry 'bout that. I should wear my specs more often,
Oh well, at least the method is there
Yeah, right ... but it's more difficult because in this case AOB is not 90 degree. It's seems quite complicated to find that
0
14 years ago
#8
(Original post by BCHL85)
Yeah, right ... but it's more difficult because in this case AOB is not 90 degree. It's seems quite complicated to find that
well you can use more trig to find the length of AB, then use the cosine rule to find the angle AOB, then use Ar = ½.OA.OB.sin(AOB).
Or you can do it directly using the lengths of the sides only.

Ar = ¼√[4b²c² - ( b² + c² - a²)²]
0
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