# P5 Hyperbola Question Watch

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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

Thanks in advance

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

Thanks in advance

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#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

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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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.

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.

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#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)

===========

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)

===========

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#5

Oh... small mistake from the beginning.

The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok

The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok

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#6

(Original post by

Oh... small mistake from the beginning.

The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok

**BCHL85**)Oh... small mistake from the beginning.

The hyperbola here is x^2 - 2.y^2 = 4 . But I think it's ok

Sorry 'bout that. I should wear my specs more often,

Oh well, at least the method is there

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#7

(Original post by

Aaargh, My eyes are dim

Sorry 'bout that. I should wear my specs more often,

Oh well, at least the method is there

**Fermat**)Aaargh, My eyes are dim

Sorry 'bout that. I should wear my specs more often,

Oh well, at least the method is there

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#8

(Original post by

Yeah, right ... but it's more difficult because in this case AOB is not 90 degree. It's seems quite complicated to find that

**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

Or you can do it directly using the lengths of the sides only.

Ar = ¼√[4b²c² - ( b² + c² - a²)²]

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