Consider the equation x^2 + xy + y^2 = 1
Find all points where the tangent line is parallel to the line y = −x.
Thank you for helping
well does that not mean that dy/dx = -1?
Using implicit differentiation,
2x + y + xy' + 2yy' = 0
y' = -(2x+y)/(x+2y)
so, you want
(2x+y)/(x+2y) = 1
2x+y = x+2y
y = x
plug that into the original equation and you get
x^2 + x^2 + x^2 = 1
x^2 = 1/3
so, at (1/√3,1/√3) and (-1/√3,-1/√3) dy/dx = -1
See from the URL below that this is true.
https://www.wolframalpha.com/input/?i=x^2+%2B+xy+%2B+y^2+%3D+1
Thank youuu so much
To find the points where the tangent line is parallel to the line y = -x, we need to find the gradient of the line y = -x. The gradient of a line is given by the coefficient of x in its equation. In this case, the gradient is -1.
Now, in order for a tangent line to be parallel to a given line, the gradients of the two lines must be equal. Therefore, the slope of the tangent line to the curve x^2 + xy + y^2 = 1 must be -1.
To find the gradient of the curve at a given point, we can take the derivative of the equation with respect to x. However, since we want to find the points where the gradient is -1, we can set up the following equation:
dy/dx = -1
We can now solve this equation simultaneously with the equation of the curve x^2 + xy + y^2 = 1.
Differentiating the curve equation implicitly with respect to x, we get:
2x + y(dx/dx) + x(dy/dx) + 2y(dy/dx) = 0
Simplifying the equation, we have:
2x + y + xy' + 2yy' = 0
We substitute -1 for dy/dx and solve for x and y:
2x + y + x(-1) + 2y(-1) = 0
2x + y - x - 2y = 0
x - y = 0
x = y
Substituting x = y back into the curve equation, we have:
x^2 + x^2 + x^2 = 1
3x^2 = 1
x^2 = 1/3
x = ±√(1/3)
Therefore, the points where the tangent line is parallel to the line y = -x are given by:
(√(1/3), √(1/3)) and (-√(1/3), -√(1/3))
So, the two points are (√(1/3), √(1/3)) and (-√(1/3), -√(1/3)).
I hope this helps! Let me know if you have any further questions.
To find the points where the tangent line is parallel to the line y = -x, we need to find the gradient (or slope) of the line y = -x and then determine which points on the given equation have the same gradient.
First, let's find the slope of the line y = -x. The general form of a line equation is y = mx + b, where m represents the slope. Comparing this with y = -x, we can see that the slope is -1.
To find the points where the tangent line is parallel, we need to differentiate the given equation implicitly with respect to x. Let's differentiate the equation x^2 + xy + y^2 = 1 with respect to x:
d/dx (x^2 + xy + y^2) = d/dx (1)
2x + (x(dy/dx) + y) + 2y(dy/dx) = 0
Next, we can rearrange the equation to solve for dy/dx:
2x + xy + y + 2y(dy/dx) = 0
2y(dy/dx) = -(2x + xy + y)
dy/dx = -(2x + xy + y) / (2y)
Now, we can equate the slope dy/dx to -1 (because the tangent line should be parallel to y = -x):
-(2x + xy + y) / (2y) = -1
To simplify further, we can multiply both sides of the equation by 2y:
2x + xy + y = -2y
Now, let's substitute y = -x into the equation:
2x + x(-x) - x = -2(-x)
2x - x^2 - x = 2x
x^2 + 3x = 0
Factoring the equation:
x(x + 3) = 0
Setting each factor equal to zero:
x = 0 or x = -3
Now, we can substitute these values of x back into the original equation to find the corresponding y-values:
For x = 0:
0^2 + 0y + y^2 = 1
y^2 = 1
y = -1 or y = 1
For x = -3:
(-3)^2 + (-3)y + y^2 = 1
9 - 3y + y^2 = 1
y^2 - 3y + 8 = 0
We can use the quadratic formula to find the y-values:
y = (-(-3) ± sqrt((-3)^2 - 4(1)(8))) / 2(1)
y = (3 ± sqrt(9 - 32)) / 2
y = (3 ± sqrt(-23)) / 2
Since the discriminant is negative, there are no real solutions for y when x = -3.
Therefore, the points where the tangent line is parallel to y = -x are (0, 1) and (0, -1).