Answer: To obtain the relation between α and β, we can eliminate t from the given equations.
(iv)
α = 1/2a(t+1/t)
β = 1/2a(t-1/t)
We can multiply these two equations to eliminate t^2:
αβ = (1/2a(t+1/t))(1/2a(t-1/t))
αβ = (1/4a^2)(t^2 - 1/t^2)
Multiplying both sides by 4a^2 gives:
4a^2αβ = t^2 - 1/t^2
Adding 1/t^2 to both sides gives:
4a^2αβ + 1/t^2 = t^2 + 1/t^2
Multiplying both sides by t^2 gives:
4a^2αβt^2 + 1 = t^4 + 1
Rearranging and simplifying gives the relation between α and β:
4a^2αβ = t^4 - 4a^2t^2 + 1
Now we can write the locus of the point as t varies:
4a^2αβ = t^4 - 4a^2t^2 + 1
This is a fourth degree equation in t, which represents a curve in the (α, β) plane. However, we can simplify it by noting that t^2 is always non-negative. Therefore, we can treat 4a^2t^2 as a constant and write:
4a^2αβ = (t^2 - 2a^2)^2 + 1 - 4a^4
This is the equation of a conic section called a hyperbola. Its center is at (0,0), its asymptotes are the lines α = ±β, and its foci are at (a√2,0) and (-a√2,0).
Explanation: