For Curves: x =1 +3t.....and...y= 2-t^2 ....find equation for line tangent to the curve at point.... t = -1?

Question

please solve using calc 2

Answer 1

When t = -1, you have (-2, 1)

t = (x - 1)/3

y = 2 - [(x-1) / 3]^2
y = 2 - ([(x-1)^2]/9)
y = 2 - (x^2 - 2x + 2)/9
y = 2 - x^2/9 + 2x/9 - 2/9
y' = -2x/9 + 2/9 or 2/9( 1 - x )
y' = 2/9(1-x)
y'(1) = 2/9(1--2) = 2/3
y - (1) = 2/3(x - -2)
y = 2/3x + 4/3 + 3/3 = 2/3x + 7/3

Answer 2

dx/dt = 3
dy/dt = - 2 t
dy/dx = (-2t) / 3
dy/ dx = 2 / 3 = m
Line passes thro' (- 2 , 1):-

y - b = m (x - a)
y - 1 = (2/3) (x + 2)
y = (2/3) x + 4 / 3 + 3 / 3
y = (2/3) x + 7 / 3