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).

x^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 )

a) To find the expression for dy/dx, we differentiate the implicit equation with respect to x:

2x - 2x(dy/dx) - 2y + 8y(dy/dx) = 0

Rearranging and factoring out dy/dx, we get:

dy/dx = (2x - 2y) / (2x - 8y)

Now, if the tangents to the curve are parallel to the x-axis, it means that dy/dx is zero. So, setting dy/dx = 0, we have:

2x - 2y = 0

Simplifying further, we get:

x = y

Therefore, the coordinates of the point where the tangents to the curve are parallel to the x-axis are (a, a), where a can be any real number.

b) To find the equation of the normal to the curve at the point (2sqrt3, sqrt3), we first need to find the slope of the tangent at that point.

Using the expression for dy/dx derived earlier:

dy/dx = (2x - 2y) / (2x - 8y)

Substituting x = 2sqrt3 and y = sqrt3, we get:

dy/dx = (2(2sqrt3) - 2(sqrt3)) / (2(2sqrt3) - 8(sqrt3))
= (4sqrt3 - 2sqrt3) / (4sqrt3 - 8sqrt3)
= 2sqrt3 / (-4sqrt3)
= -1/2

The slope of the tangent is -1/2. Since the normal to a curve is perpendicular to the tangent, the slope of the normal is the negative reciprocal of the slope of the tangent.

Therefore, the slope of the normal is -1 / (-1/2) = 2.

Using the point-slope form of the equation of a line, where the point is (2sqrt3, sqrt3) and the slope is 2, we have:

y - sqrt3 = 2(x - 2sqrt3)

Expanding and simplifying, we get:

y = 2x - 4sqrt3 + sqrt3
= 2x - 3sqrt3

Therefore, the equation of the normal to the curve at the point (2sqrt3, sqrt3) is y = 2x - 3sqrt3.

a) To find the expression for dy/dx, we can differentiate the implicit equation with respect to x using the chain rule:

d/dx(x^2 - 2xy + 4y^2) = d/dx(12)

2x - 2y(dy/dx) + 8y(dy/dx) = 0

Rearranging the terms, we get:

(dy/dx)(8y - 2x) = 2x

dy/dx = 2x / (8y - 2x)

To determine the coordinates of the point where the tangents to the curve are parallel to the x-axis, the slope (dy/dx) must be zero. Setting dy/dx = 0, we have:

2x = 0

x = 0

Substituting x = 0 back into the original equation:

0^2 - 2(0)y + 4y^2 = 12

-2y + 4y^2 = 12

2y(2y - 1) = 12

4y^2 - 2y - 12 = 0

Solving this quadratic equation, we get two possible values for y:

y = (-(-2) ± √((-2)^2 - 4(4)(-12))) / (2(4))
y = (2 ± √(4 + 192)) / 8
y = (2 ± √196) / 8
y = (2 ± 14) / 8

Therefore, the possible y-values are y = 2 and y = -12/8 = -3/2.

For y = 2, substituting this value back into the given equation to find the corresponding x-value:

x^2 - 2xy + 4y^2 = 12
x^2 - 2x(2) + 4(2)^2 = 12
x^2 - 4x + 16 = 12
x^2 - 4x + 4 = 0
(x - 2)^2 = 0

x - 2 = 0
x = 2

Therefore, one point where the tangents to the curve are parallel to the x-axis is (2, 2).

For y = -3/2, substituting this value back into the given equation to find the corresponding x-value:

x^2 - 2xy + 4y^2 = 12
x^2 - 2x(-3/2) + 4(-3/2)^2 = 12
x^2 + 3x + 9/2 = 12
2x^2 + 6x + 9 = 24
2x^2 + 6x - 15 = 0

Solving this quadratic equation, we get:

x = (-6 ± √(6^2 - 4(2)(-15))) / (2(2))
x = (-6 ± √(36 + 120)) / 4
x = (-6 ± √156) / 4

Therefore, the two possible x-values are:

x = (-6 + √156) / 4 ≈ 0.33
x = (-6 - √156) / 4 ≈ -3.08

Therefore, the points where the tangents to the curve are parallel to the x-axis are approximately (2, 2), (0.33, -3/2), and (-3.08, -3/2).

b) To find the equation of the normal to the curve at the point (2√3, √3), we need to find the slope (m) of the tangent at this point, and then find the negative reciprocal of m to get the normal slope.

Using the expression for dy/dx from part a:

dy/dx = 2x / (8y - 2x)

Substituting x = 2√3 and y = √3:

dy/dx = 2(2√3) / (8√3 - 2(2√3))
dy/dx = 4√3 / (8√3 - 4√3)
dy/dx = 4√3 / 4√3
dy/dx = 1

Therefore, the slope of the tangent at the point (2√3, √3) is 1. The negative reciprocal of 1 is -1, so the slope of the normal is -1.

Using the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) = (2√3, √3) and m = -1:

y - √3 = -1(x - 2√3)
y - √3 = -x + 2√3
y = -x + 3√3

Therefore, the equation of the normal to the curve at the point (2√3, √3) is y = -x + 3√3.

To find the expression for dy/dx in terms of x and y, we can differentiate both sides of the equation x^2 - 2xy + 4y^2 = 12 with respect to x.

Differentiating x^2 - 2xy + 4y^2 = 12 with respect to x, we get:

2x - 2y(dy/dx) + 8y(dy/dx) = 0

Simplifying this equation, we have:

2x + 6y(dy/dx) = 0

Rearranging the equation, we get:

dy/dx = -2x / (6y) = -x / (3y)

This is the expression for dy/dx in terms of x and y.

To determine the coordinates of the point where the tangents to the curve are parallel to the x-axis, we need to solve dy/dx = 0.

Setting dy/dx = -x / (3y) equal to 0, we get:

-x / (3y) = 0

Since the numerator -x is set to zero, we have:

-x = 0

From this equation, we find x = 0.

Substituting x = 0 back into the original equation x^2 - 2xy + 4y^2 = 12, we have:

0^2 - 2(0)(y) + 4y^2 = 12

Simplifying this equation, we get:

4y^2 = 12

Dividing both sides by 4, we have:

y^2 = 3

Taking the square root of both sides, we get:

y = ±√3

So, the coordinates of the points where the tangents to the curve are parallel to the x-axis are (0, √3) and (0, -√3).

Moving on to part b) to find the equation of the normal to the curve at the point (2√3, √3), we need to find the slope of the tangent at that point. We can use the expression we found for dy/dx in terms of x and y, and substitute the values of x and y for the given point.

Substituting x = 2√3 and y = √3 into the expression dy/dx = -x / (3y), we have:

dy/dx = -(2√3) / (3√3) = -2 / 3

This is the slope of the tangent at the point (2√3, √3).

Since the normal to a curve is perpendicular to the tangent at the point of contact, the slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is 3/2.

Using the point-slope form of the equation for a line, we can find the equation of the normal to the curve at the point (2√3, √3).

Using the point-slope form:

y - y1 = m(x - x1)

Substituting x1 = 2√3, y1 = √3, and m = 3/2, we have:

y - √3 = (3/2)(x - 2√3)

Expanding and simplifying, we get:

y - √3 = (3/2)x - 3√3

Rearranging this equation, we have:

(3/2)x - y = 2√3 - √3

Simplifying further, we get:

(3/2)x - y = √3

Therefore, the equation of the normal to the curve at the point (2√3, √3) is (3/2)x - y = √3.