
William W. answered 06/19/23
Experienced Tutor and Retired Engineer
If x = 2sec(θ) then we can solve the θ:
x = 2sec(θ)
x/2 = sec(θ)
θ = sec-1(x/2)
Plugging "sec-1(x/2)" in for "θ" in the other equation gives:
y = 2tan(θ)
y = 2tan(sec-1(x/2))
Remembering that sec(θ) = hypotenuse/adjacent we can draw a triangle where "x" is the hypotenuse and "2" is the adjacent:
We can now use the Pythagorean Theorem to solve for the opposite side:
22 + (opposite)2 = x2
(opposite)2 = x2 - 4
opposite = √(x2 - 4)
Then, remembering that tan(θ) = opposite/adjacent:
tan(θ) = √(x2 - 4)/2
So y = 2√(x2 - 4)/2 or y = √(x2 - 4)
Square both sides to get y2 = x2 - 4 then subtract x2 from both sides to get:
y2 - x2 = -4 then divide both sides by -4 to get:
y2/-4 - x2/-4 = 1 or x2/4 - y2/4 = 1