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

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.

User Icon for Bot Bot answered
1 year ago

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.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To find the slope of the tangent line at any point on the graph, we need to differentiate the equation xy = 1 with respect to x.

Differentiating with respect to x:
d(xy)/dx = d(1)/dx

Using the product rule:
(x * dy/dx) + (y * dx/dx) = 0

Simplifying:
x * dy/dx + y = 0

Rearranging:
dy/dx = -y/x

Now, substitute the given point (-1, 1) into the equation y = 1/x to find the y-coordinate at that point:

y = 1/(-1) = -1

Therefore, at the point (-1, 1), the y-coordinate is -1.

Now, substitute the values of x and y into the equation dy/dx = -y/x:

dy/dx = -(-1)/(-1) = -1

So, the slope of the tangent line at the point (-1, 1) is -1.

To find the equation of the tangent line, we use the point-slope form of a linear equation:

y - y1 = m(x - x1)

Substituting the values (-1, 1) and -1 for x1, y1, and m, respectively:

y - 1 = -1(x - (-1))

Simplifying:
y - 1 = -1(x + 1)
y - 1 = -x - 1

Rearranging the equation in slope-intercept form gives the equation of the tangent line passing through the point (-1, 1):

y = -x

So, the equation of the tangent line to the hyperbola xy = 1 that passes through the point (-1, 1) is y = -x.

User Icon for Explain Bot Explain Bot answered
1 year ago

To find the equations of the tangent lines to the hyperbola xy=1 that pass through the point (-1,1), we can follow these steps:

Step 1: Differentiate the equation xy=1 with respect to x to find the derivative dy/dx.
To do this, we will use implicit differentiation. Differentiating both sides of the equation with respect to x, we get:
(1 * dy/dx) + (x * d/dx[y]) = 0
Simplifying, we get:
dy/dx + x * d/dx[y] = 0

Step 2: Solve for d/dx[y] to obtain the derivative of y with respect to x.
Rearranging the equation from step 1, we get:
d/dx[y] = -dy/dx / x

Step 3: Substitute the given point (-1,1) into the equation xy=1 to find the corresponding coordinates (x,y).
Plugging in the values of x and y, we have:
(-1)(1) = 1
-1 = 1
This is not true, so (-1,1) does not lie on the hyperbola xy=1, which means there are no tangent lines through (-1,1) on the hyperbola.

Moreover, the lines tangent to the hyperbola xy=1 will not pass through the points (1,1) and (-1,-1). You may have mistaken them for the points of intersection between the hyperbola and the asymptotes. The asymptotes of the hyperbola intersect at the origin (0,0), not (1,1) and (-1,-1).

Therefore, there are no tangent lines to the hyperbola xy=1 that pass through the point (-1,1).