hard locus hyperbola qn :S (1 Viewer)

[bos1234]

PQ is a variable chord of the hyperbola

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with a constant gradient m. Show that the locus of the midpoint of PQ is the diameter

Could someone interpret this as a diagam on paint etc?

 

[Trebla]

Let P be (a sec A, b tan A) and Q be (a sec B, b tan B)
gradient of PQ = m
.: b(tan A - tan B) / a(sec A - sec B) = m

.: ma² = ab(tan A - tan B) / (sec A - sec B)


Midpoint of PQ:
x = a(sec A + sec B) / 2
y = b(tan A + tan B) / 2

b²x - ma²y = ab²(sec A + sec B) / 2 - ab²(tan A - tan B)(tan A + tan B) / 2(sec A - sec B)
= [ab²(sec A + sec B)(sec A - sec B) - ab²(tan A - tan B)(tan A + tan B)] / 2(sec A - sec B)
= ab²(sec²A - sec²B - tan²A + tan²B) / 2(sec A - sec B)
= ab²(sec²A - tan²A - (sec²B - tan²B)) / 2(sec A - sec B)
= ab²(1 - 1) / 2(sec A - sec B)
= 0 as required

 

[SeDaTeD]

I don't think he was actually asking for a solution, more an interpretation.

A diameter of any conic is a chord which passes through the centre (in this case, the origin). In this example you want to show that the midpoints of all chords PQ with gradient m lie on this particular diameter, given by that equation.

 

[bos1234]

kk i got it now

---- but look at this qn-----

The tangent to the hyperbola at a point P on the hyperbola

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meets the axes at Q and S. If OQRS is a rectangle, find the eqn of R

I got the coordinates of R(a/sec@, -b/tan@)

so x =a/sec@
y = -b/tan@

back of the book makes sec@ and tan@ as subjects then squares both simultaneous eqns and adds them

a^2/x^2 - b^2/y^2 = 1

What are they solving for?

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thanks for that worked soln. trebla

 

[SeDaTeD]

Does it mean, find the equation of the locus of R?

 

[bos1234]

ya thats right.. find the eqn of locus R..

Why did they eliminate the trigs?