Question

, Find an equation of the tangent line to the graph of x^2+3y^2=4 at the point (1,1)...
Please and thank you :)

Answer 1

I did it!

Answer 2

Good job!... Wanna let the incredibly confused in on it?

Answer 3

x^2+3y^2=4 at the point (1,1)...
2x+6yy '=0
y '=-2/6=-1/3 at point (1,1)
y-1=-(1/3)(x-1) ans or
y=-(1/3)x+ 4/3 ans