3 Find the points x y at which the curve has a vertical tang

3.) Find the points (x, y) at which the curve has a vertical tangent.

-sint = (x-10)

cost = y

so sin^2 (t) + cos^2(t) =1

(x-10)^2 + y^2 = 1

option C

x=2t

y = cos (pi*t)

t=4

then x =8 and y = 1

x\'=2

y\' = -pi*sin(pi*t)

at t=4,

y\'=0

and x\'=2

so slope = y\'/x\' = 0

line passing through (8,1) and slope of 0

1= 0 +c

c=1

so line is y=1

option b.

x=t^2 -6t

y = t^3 -12t

for vertical tengent, slope should be infinity

hence y\'/x\' = infinity

so x\' = 0

d/dx (t^2 - 6t) = 0

2t-6=0

t = 3

so x = 9-18 = -9

y = 27 - 12*3 = -9

so point (-9,-9) so option B