1. At what point does the normal line to the curve x^2 - XY + Y^2 = 3 at the point (-1,1) intersect the curve again?

2. Find the constants A, B so that if Y=A*sin X + B cos X, then Y satisfies the differential equation Y" + 2Y = 0.

3. Find the points on he graph of Y = e^x - e^2x at which that tangent line is horizontal.

Please help, I'm not sure how to do these.

#1 still does not contain the point (1,1)

#2 I don't think there is such an A and B. There is a solution involving sin(√2 x), but that's not gonna fit the requirements.

#3
y = e^x - e^2x
y' = e^x - 2e^2x
= e^x(1-2e^x)
so, y'=0 when
e^x=0 (never)
1-2e^x = 0
x = -ln2

Sure, I'd be happy to help you with these questions! Let's break them down one by one and explain how to approach each problem.

1. To find the point at which the normal line to the curve x^2 - XY + Y^2 = 3 at the point (-1,1) intersects the curve again, we need to use the concept of the normal line and its relationship to the derivative of the curve.

First, we need to find the equation of the tangent line at the given point (-1,1). To do this, we take the derivative of the curve equation with respect to x:

d/dx(x^2 - XY + Y^2) = 0

This gives us the slope of the tangent line at any point on the curve. Next, we substitute the x and y values of the given point (-1,1) into the derivative equation to find the slope of the tangent line at (-1,1).

Once we have the slope of the tangent line, we can find the equation of the normal line by taking the negative reciprocal of the tangent line's slope. This gives us the slope of the normal line.

Next, we use the equation of the normal line and the given point (-1,1) to find the equation of the normal line.

Finally, we solve the system of equations formed by the curve and the normal line to find the point of intersection.

2. To find the constants A and B so that Y = A*sin(X) + B*cos(X) satisfies the differential equation Y'' + 2Y = 0, we need to substitute Y into the differential equation and solve for the constants.

First, take the second derivative of Y with respect to X. Then substitute Y, Y', and Y'' into the differential equation. This should give you an equation involving A and B.

Next, we solve the resulting equation with A and B as variables. This can typically be done by combining like terms and applying algebraic techniques, such as factoring or manipulating equations.

By solving the equation, you will find the values of A and B that satisfy the given differential equation.

3. To find the points on the graph of Y = e^X - e^(2X) at which the tangent line is horizontal, we need to find the points of the graph where the first derivative is zero.

First, take the derivative of Y with respect to X. Then set the derivative equal to zero and solve for X. This will give you the x-values of the points where the tangent line is horizontal.

Once you have the x-values, substitute them back into the original equation Y = e^X - e^(2X) to find the corresponding y-values.

The resulting x and y-values will be the points on the graph where the tangent line is horizontal.

I hope this explanation helps you understand how to approach and solve these questions! If you have any further questions, feel free to ask.