Finding Horizontal Tangents from Implicit Equations

What Makes a Tangent Horizontal?

A tangent line is horizontal when its slope is zero. For a curve described by an equation relating \(x\) and \(y\), implicit differentiation gives a way to calculate the slope without first solving the equation explicitly for \(y\).

The key is to treat a horizontal tangent as a point-finding problem with several checks. Setting the numerator of the implicit slope formula to zero finds possible points, but those candidates must also lie on the original curve, and the denominator must be nonzero so the slope formula applies.

The Implicit Slope Formula

Suppose a curve is given by \(F(x,y)=0\), where \(F\) has continuous first partial derivatives near a point on the curve. At a point where \(F_y\ne 0\), the implicit function theorem guarantees that the equation describes \(y\) locally as a differentiable function of \(x\). Differentiating \(F(x,y(x))=0\) by the chain rule gives

Solving for the derivative is valid only when \(F_y(x,y)\ne 0\):

Here, \(F_x\) means the partial derivative with respect to \(x\), holding \(y\) constant, and \(F_y\) means the partial derivative with respect to \(y\), holding \(x\) constant. The formula expresses the slope at a point of the curve; it does not say that every pair \((x,y)\) is on the curve.

Why the Numerator Must Be Zero

At a point where the implicit slope formula applies, the tangent is horizontal exactly when its slope equals zero. A fraction with nonzero denominator is zero exactly when its numerator is zero. Thus the slope formula gives the condition \(F_x=0\), provided \(F_y\ne 0\). But the point must also satisfy the original relation \(F(x,y)=0\). These conditions work together, not separately.

The condition \(F(x,y)=0\) confirms that the candidate lies on the curve. The condition \(F_x(x,y)=0\) makes the slope zero. The condition \(F_y(x,y)\ne 0\) allows us to solve the differentiated equation for \(\frac{dy}{dx}\) and ensures the local graph has a well-defined derivative there.

A Reliable Procedure

Worked Examples

Common Mistake: Stopping After Setting the Numerator to Zero

A common error is to solve \(F_x=0\) and immediately report the resulting coordinates as horizontal tangent points. The numerator condition only produces candidates. It does not prove that those coordinates are real points on the curve.

For a concrete example, consider \(x^2+y^2=-1\). Differentiation gives \(2x+2y\frac{dy}{dx}=0\), so a candidate for a horizontal tangent must have \(x=0\). But substituting \(x=0\) into the original relation gives \(y^2=-1\), which has no real solution. There are no real candidate points on this relation, so the calculation does not produce a horizontal tangent.

Why the Checks Matter

Implicit differentiation is useful precisely because many curves are easier to describe by a relation than by an explicit formula \(y=f(x)\). The horizontal-tangent test uses that relation directly: it turns a geometric question about a tangent into algebraic conditions involving the relation and its partial derivatives.

However, each condition answers a different question. The original equation asks whether the candidate is on the curve. The numerator asks whether the slope is zero. The denominator asks whether the slope formula is valid at that point. Omitting any one of these checks can lead to a false conclusion.

When the checks pass at \((a,b)\), the derivative is zero there, so the tangent line has equation \(y-b=0(x-a)\), which simplifies to \(y=b\). This gives both the tangent point and the horizontal tangent line.