Tuesday

July 29, 2014

July 29, 2014

Posted by **Matteo** on Monday, March 7, 2011 at 3:53pm.

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).

- math -
**Reiny**, Monday, March 7, 2011 at 5:29pmx^2-2xy+4y^2=12

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 )

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