Find the equation of the tangent line at the given point on

Find the equation of the tangent line at the given point on the curve. x^2 3y^2 = 13; (1, 2) y = -1/3 x + 7/3 y = 1/6 x + 11/6 y = - -1/6 x Write the word or phrase that best completes each statement answers the question. Assume x and y are functions of t. Evaluate dy/dt. x^3 = 12y^5 -4;dx/dt = 6, y = 1

Solution

14)
x^2 + 3y^2 = 13
differentiate with respect to x,
2x + 6y*dy/dx = 0
dy/dx = -2x/6y
= -x/3y

at (1,2)
slope = dy/dx = -1/6

equation of tangent:
(y-yo) = slope * (x-xo)
(y-2) = (-1/6)*(x-2)
6y-12 = -x+2
6y = -x+14
y = (-1/6)*x + 7/3

i am allowed to answer only 1 question at a time