math
posted by Matteo .
A curve has implicit equation x^22xy+4y^2=12
a)find the expression for dy/dx in terms of y and x. hence determine the coordinates of the point where the tangents to the curve are parallel to the xaxis.
b)Find the equation of the normal to the curve at the point (2sqrt3,sqrt3).

x^22xy+4y^2=12
2x  2x dy/dx  2y + 8y^ dy/dx = 0
dy/dx(8y^2  2x) = 2y  2x
dy/dx = (yx)/(4y^2x)
to be horizontal to the xaxis, the slope must be zero, that is, dy/dx = 0
so yx = 0
y = x
plug that back into the original
x^2  2x^2 + 4x^2 = 12
3x^2 = 12
x = ±2
two points: (2,2) and (2,2)
b) plug the x and y values into dy/dx to get the slope
take the negative reciprocal to get the slope of the normal
use your grade 9 method to find the equation, now that you have the slope m and a given point.
(I had y = 2x + 5√3 )
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