Your calculations are fine, haebinpark.
Maybe the problem was with the graphing.
The slopes of the tangents are 0.5 and -0.5,
which you will get if you simplify a little more
the question is
use implicit differentiation to find the slope of tangent at x = sqrt 2 for x^2 + 4y^2 = 4
what i did was
dx^2 / dx = 2x
d4y^2 / dx = 8yy'
d4/dx = 0
2x + 8yy' = 0
8yy' = -2x
y'= -2x/8y
= -x/4y
since x = sqrt 2
(sqrt 2)^2 + 4y^2 = 4
2+4y^2 = 4
4y^2 = 2
y^2 = 2/4 = 1/2
y= +/- sqrt 1/2
since y' = -x / 4y and x=sqrt 2, and y = +/- sqrt 1/2
y'= - sqrt2 / 4(sqrt 1/2)
and
y'= - sqrt2 / 4(- sqrt 1/2) = sqrt2 / 4(sqrt 1/2)
so i got slope of tangent = +/- sqrt2 / 4(sqrt 1/2)
but i drew graph on the calculator and seems like i did something wrong
but i can't find where i did go wrong...
does anyone see what i did wrong?