AP® Calculus AB · Unit 3 · Topic 3.2

Tangent Lines to Implicit Curves

Once you have dy/dx from implicit differentiation, you can write a tangent line at any point by plugging in coordinates for the slope. Horizontal tangents occur where the numerator of dy/dx is zero; vertical tangents occur where the denominator is zero.

An implicit curve like x2 + xy + y2 = 7 is not a function, so it can loop back on itself and have tangent lines pointing in every direction, including straight up. A regular function y = f(x) can never have a vertical tangent, because that would mean two y-values for one x-value. But an implicit curve is allowed to do that, which is exactly why it can have a point where the tangent is vertical: the curve momentarily behaves like it has infinite slope, which is the signature of a function failing to be a function at that spot.

The method is always the same. Differentiate both sides of the equation with respect to x, treating y as a hidden function of x and using the chain rule every time you touch a y, then solve the resulting equation for dy/dx. The result is almost always a fraction built from x and y. That fraction tells you everything: plug in a specific point to get the slope there for a tangent line equation, set the numerator equal to 0 to find horizontal tangents (slope zero), and set the denominator equal to 0 to find vertical tangents (slope undefined). In every case you still need to check that the point actually lies on the original curve.

This topic is the payoff for learning implicit differentiation in the previous lesson: now that you can compute dy/dx without solving for y, you can do all the same tangent-line and shape analysis you did for ordinary functions. It sets up later work on related rates and curve analysis, where implicit relationships between changing quantities are the norm rather than the exception.

Key terms

implicit differentiation
Differentiating both sides of an equation in x and y, applying the chain rule to every y term.
horizontal tangent
A tangent line with slope 0, found where the numerator of dy/dx equals 0.
vertical tangent
A tangent line with undefined slope, found where the denominator of dy/dx equals 0.
point-slope form
The equation y = y1 + m(x - x1) for a line through (x1, y1) with slope m.

Watch it explained

  1. How it worksHow do you locate the points on an implicit curve with a horizontal tangent line?
  2. Worked exampleFind the tangent line to x2 + xy + y2 = 7 at the point (1, 2).
  3. Worked exampleFind every point on the ellipse x2 + 4y2 = 36 with a vertical tangent line.

Check yourself

Your turn1 of 3
A curve satisfies x2 + xy + y2 = C for some constant C. Find the tangent line to the curve at the point (2, 2), giving y as a function of x.

End of topic.