Equations of the tangent lines to hyperbola xy=1 that pass through point (-1,1)

Question

I know the graph of y=1/x but not sure about the tangent lines at given point. Are the lines tangent to (1,1) and (-1,-1)?

You're correct, the graph is y=1/x, so find y' to get the slope at any point on the graph (except the origin). I only found one tangent line through (-1,1), I'm not sure what other point the question has in mind.

Answer 1

To find the equation of the tangent lines to the hyperbola xy = 1 that pass through the point (-1, 1), we first need to find the derivative of the hyperbola. Using implicit differentiation, we have:

x(dy/dx) + y = 0

Solving for dy/dx, we get:

dy/dx = -y/x

Now, let's consider a tangent line that passes through the point (-1, 1). We need to find a point (x_1, y_1) on the hyperbola such that the slope between (-1, 1) and (x_1, y_1) equals the value of the derivative at (x_1, y_1).

Using the point-slope form of a line, we have:

(y - 1) / (x + 1) = -(y_1) / (x_1)

But since (x_1, y_1) is a point on the hyperbola, we have x_1 * y_1 = 1.

Now we have a system of equations:

(y - 1) / (x + 1) = -(y_1) / (x_1)
x_1 * y_1 = 1

Solve this system to find (x_1, y_1). We'll first express y_1 in terms of x_1:

y_1 = 1 / x_1

Now substitute this into the first equation:

(y - 1) / (x + 1) = -(1 / x_1) / (x_1)

Multiply both sides by x_1(x + 1) to eliminate the denominators:

x_1(y - 1) = -(x + 1)

Now substitute y = 1/x back in:

x_1(1/x - 1) = -(x + 1)

Simplify and solve for x:

x^2 - x_1^2 = -x_1^2 - x_1

x^2 = -x_1

But since x^2 is always non-negative, there is no solution for x_1, and therefore no tangent line that passes through the point (-1, 1). The point (-1, 1) does not lie on the hyperbola, so there is no tangent line from the point to the hyperbola.