Parametric equation x = (e ^ t) Sint y = (e ^ t) cost t = 90 degree tangent equation

Parametric equation x = (e ^ t) Sint y = (e ^ t) cost t = 90 degree tangent equation

x't=(e^t)(sint+cost),y't==(e^t)(cost-sint)
So the slope k = y't / X't = (cost Sint) / (Sint + cost)
t=∏/2,
Tangent point x = e ^ (Π / 2), y = 0
k=-1,
So the tangent equation y = - (x-e ^ (Π / 2)), is y = - x + e ^ (Π / 2)

Parameter equation x = t + 1 / T-1, y = 2T / T ^ 3-1
How to change ordinary equation

X-1 = (T + 1) / (t-1) - 1 = 2 / (t-1) T-1 = 2 / (x-1) t = (x + 1) / (x-1) T ^ 2 + T + 1 = (x + 1) ^ 2 / (x-1) ^ 2 + (x + 1) / (x-1) + 1 = (3x ^ 2 + 1) / (x-1) ^ 2, so T ^ 3-1 = (t-1) (T ^ 2 + T + 1) = [2 / (x-1)] [(3x ^ 2 + 1) / (x-1) ^ 2] so y = 2 [(x + 1) / [2 / (x-1)] [(3x ^ 2 + 1) / (x-1) = [2 / (x-1)]

It is known that the parameter equation of the straight line is: x = - 1 + T, y = - 2-2t (t is the parameter), it intersects with the ellipse 4x ^ 2 / 9 + y ^ 2 / 9 = 1 at a, B, and calculates the length of ab

Substituting X and y of linear parametric equation into elliptic equation 4 * (- 1 + T) ^ 2 + (- 2-2 * t) ^ 2 = 9, we get T ^ 2 = 5 / 8t = + (-) 0.790569415042095, two points a (- 0.209430584957905, - 3.581138830084190) B (- 1.790569415042095, - 0.418861169915810), the distance is 1.62

The parameter equation of the line L is x = t + 1, y = T-1 (t is the parameter), P (x, y) is the point on the ellipse x ^ 2 / 4 + y ^ 2 = 1. Find the maximum distance between the point P and the line L

Solution
The parameter equation of ∵ line L is x = t + 1, y = T-1
The linear equation is x-y-2 = 0
Let the point P coordinate be (2cos θ, sin θ), then the distance from point P to line L is (θ∈ [0,2 π])
|2cos θ - sin θ - 2 | / √ 2 = | √ 5cos (θ + ξ) - 2 | / √ 2 (ξ is the auxiliary angle)
When cos (θ + ξ) = - 1, the maximum distance is (√ 5 + 2) / 2 = (√ 10 + 2 √ 2) / 2
The maximum distance from P to line L is (√ 10 + 2 √ 2) / 2