umm i don't remember how to find the slope for this problem.
use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)
e^y + x^2 - 3y^2=5
so i know the tangent formula is (y-y1) = m (x-x1)
umm i don't remember how to find the slope for this problem.
use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)
e^y + x^2 - 3y^2=5
so i know the tangent formula is (y-y1) = m (x-x1)
Well,
you have x1 = 2 , y1 = 0 ..
By differentiating both sides w.r.t. x, we got:
$\displaystyle y' e^y + 2x - 6yy' = 0 $
solve for y' , then evaluate it at (2,0), i.e. when x=2 and y=0 to get the value of m ..
then substitute x1,y1 and m the tangent line's formula ..
Please, note that its called "tangent line's formula" not "tangent formula" ..