There are lines that pass through the point (-1, 3) and are tangent to the curve xy = 1. Find all their slopes.
A) -1 and -9
B) -1 and 9
C) 1 and 9
D) 1 and -9
E) none of the above

Question

Quizplus · Accepted Answer

The equation of the curve is xy = 1. The derivative, which gives the slope of the tangent at any point (x, y) on the curve, is found by implicit differentiation: d(xy)/dx = d(1)/dx, yielding y + x(dy/dx) = 0. Solving for dy/dx gives the slope of the tangent line at any point as -y/x. For a line to be tangent to the curve and pass through the point (-1, 3), it must satisfy the equation of the tangent line at some point (x, y) on the curve. Setting up the equation of a line through (-1, 3) with slope -y/x, we get y - 3 = (-y/x)(x + 1). Substituting xy = 1 into this equation allows us to solve for the values of y that make the line tangent to the curve, leading to the slopes of -1 and -9.

There Are Lines That Pass Through the Point (-1, 3)