posted by Matteo .
A curve has implicit equation x^2-2xy+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 x-axis.
b)Find the equation of the normal to the curve at the point (2sqrt3,sqrt3).
2x - 2x dy/dx - 2y + 8y^ dy/dx = 0
dy/dx(8y^2 - 2x) = 2y - 2x
dy/dx = (y-x)/(4y^2-x)
to be horizontal to the x-axis, the slope must be zero, that is, dy/dx = 0
so y-x = 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 )