For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.
For the curve given by 4x^2 + y^2 = 48 + 2xy, find the positive y-coordinate given that the x-coordinate is 2.
if the line is horizontal, then dx/dy==0
8xdx+2ydy=2xdy+2ydx
dy/dx (2y-2x)=2y-8x
or 2y=8x or y=4x
but if this is true at x=2, then y must be 8.
In the original equation
4(4)+8^2 =? 48+2(2*8)
80=? 80, which is true, so at x=2 dy/dx is zero at the point of tangency
To find the point P with x-coordinate 2 where the tangent line is horizontal, we need to find the value of y at that point.
Step 1: Differentiate the equation of the curve with respect to x to find the slope of the tangent line.
Differentiating both sides of the equation 4x^2 + y^2 = 48 + 2xy with respect to x, we get:
8x + 2yy' = 2y + 2xy'
Step 2: Simplify the derivative equation.
Rearranging the terms, we have:
8x - 2y = 2xy' - 2yy'
Step 3: Substitute the x-coordinate of P into the equation.
Since we are looking for a point P with x-coordinate 2, we substitute x = 2 into the equation:
8(2) - 2y = 2(2)y' - 2y^2
Simplifying further, we get:
16 - 2y = 4y' - 2y^2
Step 4: Set the derivative equal to 0 to find the point where the tangent line is horizontal.
Since the tangent line is horizontal when its slope is zero, we set the derivative equal to 0:
4y' - 2y^2 = 16 - 2y
Step 5: Solve the equation for y.
Rearranging the terms, we have:
2y^2 - 4y + 16 = 0
Step 6: Solve the quadratic equation.
Using the quadratic formula, we have:
y = [4 ± sqrt((-4)^2 - 4(2)(16))] / (2(2))
y = [4 ± sqrt(16 - 128)] / 4
y = [4 ± sqrt(-112)] / 4
Since the square root of a negative number is not defined in the set of real numbers, there are no real solutions for y. Therefore, there is no point P with x-coordinate 2 where the tangent line is horizontal for the given curve.
To show that there is a point P with x-coordinate 2 at which the tangent line to the curve is horizontal, we need to find the derivative of the curve with respect to x and then evaluate it at x = 2.
First, let's differentiate both sides of the equation 4x^2 + y^2 = 48 + 2xy with respect to x. This will give us the derivative of the curve with respect to x.
Differentiating the left side:
d/dx (4x^2) = 8x
d/dx (y^2) = 2y * dy/dx (using the chain rule)
Differentiating the right side:
d/dx (48 + 2xy) = 2y + 2x * dy/dx
Setting these derivatives equal to each other, we get:
8x + 2y * dy/dx = 2y + 2x * dy/dx
Next, we want to find dy/dx, which represents the slope of the tangent line to the curve. Rearranging the equation, we can isolate dy/dx:
8x - 2x * dy/dx = 2y - 2y * dy/dx
(8x - 2x) * dy/dx = 2y - 2y
6x * dy/dx = 0
Now, we can solve for dy/dx:
dy/dx = 0 / (6x)
dy/dx = 0
Since dy/dx = 0, this means that the tangent line to the curve is horizontal at the point P with x-coordinate 2.
Therefore, we have shown that there is a point P with x-coordinate 2 at which the tangent line to the curve is horizontal.