Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?
where is dy/dx = + or - oo ????
2 x dx - x dy - y dx - 2 y dy = 0
(2x-y) dx - (2y+x)dy = 0
dy/dx = (2x-y)/(2y+x)
when is 2y+x = 0?
when x = -2y
(-2y)^2 + 2 y^2 -y^2 = 5
3 y^2 = 5
y = +/- sqrt (5/3)
x - -/+ 2 sqrt(5/3)
check my arithmetic. I did that fast.
To find the points on the curve where the tangent lines are vertical, we need to find the derivative of the curve with respect to x and set it equal to zero. The derivative of the curve x^2 - xy - y^2 = 5 can be calculated using implicit differentiation.
Let's begin by differentiating both sides of the equation with respect to x:
d/dx(x^2 - xy - y^2) = d/dx(5)
2x - (x(dy/dx) + y) - 2y(dy/dx) = 0
Now we can solve for dy/dx by moving all the terms involving dy/dx to one side:
2x - xy - 2y(dy/dx) + y(dy/dx) = 0
Grouping the terms containing dy/dx together:
(2x - 2y + y)dy/dx = xy - 2x
Simplifying further:
(2x - y)dy/dx = (x - 2x)
(2x - y)dy/dx = -x
Finally, we can find dy/dx:
dy/dx = -x / (2x - y)
To find the points on the curve where the tangent lines are vertical, we set the derivative dy/dx equal to infinity. This occurs when the denominator (2x - y) equals zero:
2x - y = 0
Rearranging the equation:
y = 2x
Substituting this value of y back into the original curve equation:
x^2 - x(2x) - (2x)^2 = 5
Simplifying:
x^2 - 2x^2 - 4x^2 = 5
-5x^2 = 5
Dividing both sides by -5:
x^2 = -1
Since the square of any real number cannot be negative, there are no real solutions for x. Therefore, there are no points on the curve x^2 - xy - y^2 = 5 where the tangent lines are vertical.
To determine which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines, we need to find where the derivative of the curve with respect to x is equal to infinity.
Here's how you can do it step by step:
Step 1: Differentiate the equation with respect to x.
Differentiating the equation x^2 - xy - y^2 = 5 with respect to x using the rules of differentiation, we get:
2x - (y + xy') - 2yy' = 0
Step 2: Solve for y'.
Rearrange the equation to solve for y':
2x - xy' - y - 2yy' = 0
Combining like terms, we get:
(2x - y) - y' (x + 2y) = 0
Now solve for y':
y' = (2x - y) / (x + 2y)
Step 3: Find where y' approaches infinity.
To find where y' approaches infinity, we need to set the denominator equal to zero and solve for the corresponding x and y values. So, we have:
x + 2y = 0
Step 4: Solve for x and y.
Rearrange the equation x + 2y = 0 to solve for x:
x = -2y
Substitute this x value into the original equation x^2 - xy - y^2 = 5:
(-2y)^2 - (-2y)y - y^2 = 5
4y^2 + 2y^2 - y^2 = 5
5y^2 = 5
y^2 = 1
y = ±1
Step 5: Find the corresponding x values.
Substitute the y values into x = -2y:
For y = 1:
x = -2(1) = -2
For y = -1:
x = -2(-1) = 2
Therefore, the points on the curve of x^2 - xy - y^2 = 5 that have vertical tangent lines are (-2, 1) and (2, -1).