Explanation:
We have:
x = a sin 2t
Differentiating with respect to t, we get:
dx/dt = 2a cos 2t
Next, we have:
y = a(cos 2t + log tan t)
Differentiating with respect to t, we get:
dy/dt = -2a sin 2t + (1/tan t)(1/ln 10)
Using the identity:
sin^2 t + cos^2 t = 1
We have:
sin 2t = 2sin t cos t
And:
cos 2t = cos^2 t - sin^2 t
cos 2t = 2cos^2 t - 1
Using these identities, we can rewrite dx/dt and dy/dt in terms of x and y:
dx/dt = 2a sqrt(1 - x^2/a^2)
dy/dt = -2a sqrt(1 - x^2/a^2) + (1/ln 10)(y - a cos 2t)
Therefore, we have:
dx/dy = dx/dt ÷ dy/dt
Substituting the expressions for dx/dt and dy/dt, we get:
dx/dy = (2a sqrt(1 - x^2/a^2)) / (-2a sqrt(1 - x^2/a^2) + (1/ln 10)(y - a cos 2t))
Simplifying, we get:
dx/dy = (-2 sqrt(1 - x^2/a^2)) / (2 sqrt(1 - x^2/a^2) - (1/ln 10)(y - a cos 2t))