Find the point(s) (if any) of horizontal tangent lines: x^2+xy+y^2=6
2x + y + xy' + 2yy' = 0
y'(x+2y) = -(2x+y)
y' = -(2x+y)/(x+2y)
So, y'=0 when 2x+y = 0
So, if y = -2x,
x^2-2x^2+4x^2 = 6
3x^2 = 6
x = √2
So, y = -2√2
At (√2,-2√2) and (-√2,2√2) the tangent is horizontal.
Check:
http://www.wolframalpha.com/input/?i=plot+x^2%2Bxy%2By^2%3D6+and+y%3D2%E2%88%9A2+and+y%3D-2%E2%88%9A2
Sure, let's find the point(s) of horizontal tangent lines on the curve x^2 + xy + y^2 = 6. To do this, we'll need to find the derivative with respect to x and set it equal to zero.
Taking the derivative implicitly, we get:
2x + (1y + x * dy/dx) + 2y * dy/dx = 0
Now, we know we're looking for horizontal tangent lines, which means the slope of the tangent line (dy/dx) should be zero. Plugging this into our equation, we get:
2x + y = 0
This equation represents the slope of the tangent line at any given point on the curve. To find the points of horizontal tangents, we need to solve this equation along with the original equation x^2 + xy + y^2 = 6.
Solving these two equations simultaneously, we can substitute 2x + y = 0 into x^2 + xy + y^2 = 6:
x^2 + x(-2x) + (-2x)^2 = 6
x^2 - 2x^2 + 4x^2 = 6
3x^2 = 6
x^2 = 2
Solving for x, we get x = ±√2.
Now, plugging these x-values back into the equation 2x + y = 0 to find y-values, we get:
For x = √2: 2√2 + y = 0 => y = -2√2
For x = -√2: -2√2 + y = 0 => y = 2√2
So, the two points at which the curve has horizontal tangent lines are (√2, -2√2) and (-√2, 2√2).
Hope that clears things up!
To find the point(s) of horizontal tangent lines, we need to take the derivative of the given equation with respect to y and set it equal to zero. Let's start by finding this derivative.
Differentiating both sides of the equation x^2 + xy + y^2 = 6 with respect to y, we obtain:
2xy + 2yy' + 2y = 0
Simplifying this equation, we get:
2(xy + yy' + y) = 0
xy + yy' + y = 0
Now, to find the points of horizontal tangents, we need to solve this equation for y' by setting the derivative equal to zero:
yy' = -xy - y
To make it simpler, we can write the equation as:
y' = (-xy - y) / y
Now, we can set y' equal to zero and solve for y:
0 = (-xy - y) / y
Multiplying both sides of the equation by y, we get:
0 = -xy - y
Bringing all the terms to one side, we have:
xy + y = 0
Factoring out y, we get:
y(x + 1) = 0
This equation has two possible solutions:
1. y = 0
2. x + 1 = 0, which gives x = -1
So, the points of horizontal tangents for the given equation x^2 + xy + y^2 = 6 are:
1. (x, y) = (0, 0)
2. (x, y) = (-1, 0)
To find the point(s) of horizontal tangent lines, we need to find the values of x and y that satisfy the equation for the horizontal tangent line. The equation of a horizontal line can be expressed as y = k, where k is a constant.
1. Differentiate both sides of the given equation with respect to x:
d/dx (x^2 + xy + y^2) = d/dx (6)
2x + (x)(dy/dx) + 2y(dy/dx) = 0
2. To find the slope of the tangent line (dy/dx), let's rearrange the equation:
(x)(dy/dx) + 2y(dy/dx) = -2x
dy/dx(x + 2y) = -2x
dy/dx = -2x / (x + 2y)
3. Since we are looking for the points where dy/dx = 0 (horizontal tangent lines), we set the numerator equal to zero:
-2x = 0
x = 0
4. Now, substitute x = 0 back into the original equation:
(0)^2 + (0)(y) + y^2 = 6
y^2 = 6
y = ±√6
Therefore, there are two points that correspond to horizontal tangent lines: (0, √6) and (0, -√6). These points satisfy the condition that the derivative with respect to x, dy/dx, is equal to zero.