x^2-x^2y=y^2-1
find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.
sub x = 2 into your equation to get
4 - 4y = y^2 - 1
0 = y^2 + 4y - 5
(y+5)(y-1) = 0
so when x=2, y =-5 or when x=2, y = 1
so the points are (2,-5) and (2,1)
check them by subbing them back into the original equation, they work
I found the derivative of your equation implicitly and got
dy/dx = (2x+2xy)/(2y = x^2)
for a horizontal tangent this has to be zero, which means the numerator should be zero
subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)
To find all the points on the curve where x = 2, we can substitute x = 2 in the given equation and solve for y.
Start with the equation: x^2 - x^2y = y^2 - 1
Substitute x = 2:
2^2 - 2^2y = y^2 - 1
4 - 4y = y^2 - 1
Rearrange the equation to get it in the form of a quadratic equation:
y^2 + 4y - 5 = 0
Factor the quadratic equation:
(y - 1)(y + 5) = 0
Setting each factor equal to zero, we get two possible values of y:
y - 1 = 0 --> y = 1
y + 5 = 0 --> y = -5
Therefore, there are two points on the curve where x = 2: (2, 1) and (2, -5).
To show that there is a horizontal tangent to the curve at one of the points (2, -5), we need to find the derivative of the curve and evaluate it at that point.
Differentiating the given equation implicitly with respect to x, we get:
2x - (2xy + x^2(dy/dx)) = 2y(dy/dx)
Since we are interested in the horizontal tangent, which means the slope (dy/dx) is zero, we can set dy/dx = 0.
Substituting x = 2 and y = -5 into the derived equation and setting dy/dx = 0, we have:
2(2) - (2(2)(0) + 2^2(dy/dx)) = 2(-5)(0)
4 - 4(dy/dx) = 0
-4(dy/dx) = -4
dy/dx = -1
Thus, at the point (2, -5), there is a horizontal tangent to the curve.