synkk said:show that the locus of the point [itex] \left(\dfrac{a(cosh\theta + 1)}{2cosh\theta},\dfrac{b(cosh\theta - 1)}{2sinh\theta}\right) [/itex]
has equation x(4y^2 + b^2) = ab^2
working: http://gyazo.com/4c96af128d0293bce7f18029c2f54b0d
where have I gone wrong :(
haruspex said:Your 5th and 6th lines each have the form <some expression> = 1. The second should have something other than 1 on the right hand side.
Mark44 said:I have to say that the image of your work is very poor quality. IMO, you are lucky that haruspex took the time to decipher your hard-to-read work. I'm sure that many others wouldn't bother.
You're almost there. Just need to spot a factor.synkk said:[itex] -b^2a^2 + 2ab^2x - b^2x^2 = 4y^2x^2 - 4ay^2x [/itex]
not sure where to go from here
haruspex said:You're almost there. Just need to spot a factor.
A locus is a set of points that satisfy a particular geometric condition. For example, the locus of points equidistant from a given point is a circle.
Hyperbolic functions are a group of mathematical functions that are analogs of the trigonometric functions. They are defined in terms of exponential functions and have properties similar to the trigonometric functions.
The hyperbolic functions can be used to define the locus of points that satisfy certain conditions. For example, the locus of points equidistant from two given points can be described using hyperbolic functions.
The Hyperbolic Locus Theorem states that the locus of points equidistant from two given points is a hyperbola. This theorem is a fundamental concept in hyperbolic geometry.
Locus and hyperbolic functions have many applications in mathematics and engineering. They are used in the design of bridges and other structures, as well as in the analysis of electromagnetic fields and control systems. They also have applications in physics, such as in the study of special relativity.