Question

Find an equation of the tangent line to the graph of x^2+3y^2=4 at the point (1,1)
A. y+1=-1/3(x+1)
B. y-1=-x/3y(x-1)
C. x+3y=2
D. y-1=-1/3(x-1)
E. None of these

Answer 1

dy/dx = slope = m

for this ellipse

2 x dx/dx + 6 y dy/dx = 0

at(1,1)
2(1)+6(1)dy/dx = 0
dy/dx = m = -1/3
so
y = -(1/3) x + b
at (1,1)
1 = -1/3) + b
b = 4/3
so
y = -(1/3) x + 4/3
or
3 y - -x + 4