Question

find the locus of the point x=a/2(t+1/t) y=a/2(t-1/t) t is parameter

Answer 1

Ok right now I don't have a pen and a book in my hand and I m sry for that
But the process is really simple
Take a/2 on left side for both x and y
I mean
2x/a=t+1/t(equation 1)
2y/a=t-1/t(equation 2)
squaring equation 1
4x^2/a^2=t^2+2+1/t^2(equation 3)
4y^2/a^2=t^2-2+1/t^2(equation 4)
subtracting equation 3 and 4
we get
4x^2/a^2-4y^2/a^2=4
dividing throughout by 4
x^2/a^2-y^2/a^2=1
The locus is therefore hyperbola
HOPE it helps

Answer 2

Answer:

x^2-y^2=a^2

Step-by-step explanation:

x=a/2(t+1/t) , y=a/2(t+1/t)

2x/a=t+1/t , 2y/at+1/t

(2x/a)^2-(2y/a)^2=(t+1/t)^2-(t+1/t)^2

4x^2/a^2-4y^2/a^2=4t(1/t)

x^2-y^2=a^2