P5 Co-ordinate Geography (Ex 4c)Watch

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Thread starter 14 years ago
#1
Hi,

Having trouble with the following Qs

7c) Find a cartesian equation of the curve given parametrically as:
x = t+1/t
y = t-1/t

How do you do this one? I completed the square to make t the subject and substituted but the final equation looks very messy.

8) Prove that the triangle formed byt the asymptotes of the curve with equation x^2 - 2y^2 = 4 and any tangent to the curve is of constant area.

I worked out the asymptotes equations and the general tangent equation. I also have the angle between the 2 asymptotes and the points where the tangent crosses the asymptotes.

I guess you have to use Area = ABsinC. Problem is, its all getting messy.

Heres what I have:
theta (angle between tangents) = 2arctan (sqrt2/2)
equation of tangent: x(x1) - 2y(y1) = 4 At P(x1,y1)
points of intersection: x(x1) +- x(y1)sqrt2 = 4

Any ideas?
0
14 years ago
#2
For 7, we add two euations: x+y=2t, then substract them: x-y=2/y. After that we multiply them: x^2-y^2=4.

For 8, you need to find all the intersections, then you could see from the graph, the area of the big triangle consists of two small tiangles(one above x-axis, one below x-axis). We need to find out the coordinations at where the tangent meets x-axis, so you can use the coordinations (in terms of x and y) of the three intersections to work out the two small areas. Finally by using the fact of the original equation, all the x and y get cancelled, only a constant is left.
0
14 years ago
#3
(Original post by Womble548)
7c) Find a cartesian equation of the curve given parametrically as:
x = t+1/t
y = t-1/t
x = t + 1/t ---> x^2 = (t + 1/t)(t + 1/t) = t^2 + 1/t^2 + 2
y = t - 1/t ---> y^2 = (t - 1/t)(t - 1/t) = t^2 + 1/t^2 - 2

Cartesian Equation Of Curve Is Found By Eliminating t:
---> x^2 - y^2 = (t^2 + 1/t^2 + 2) - (t^2 + 1/t^2 - 2)
---> x^2 - y^2 = 4
0
Thread starter 14 years ago
#4
Thanks, managed to do the first one. Although the second one is still bugging me 0
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