Why Partial Derivatives Identify Tangent Direction
An implicit equation \(F(x,y)=0\) describes a curve without necessarily expressing \(y\) explicitly as a function of \(x\). At a point on the curve, the partial derivatives \(F_x\) and \(F_y\) reveal whether the local tangent is horizontal or vertical, provided the point is regular.
The key is to distinguish a candidate from a verified tangent. A point must first satisfy the curve equation. Then the relevant partial derivative must vanish, while the other must be nonzero. That nonzero condition ensures that the curve has a well-behaved local branch through the point.
The Classification Rule
Assume \(F\) has continuous first partial derivatives near \((a,b)\), and \(F(a,b)=0\). If \(F_y(a,b)\ne 0\), the implicit function theorem gives a local branch \(y=y(x)\). Differentiating the identity \(F(x,y(x))=0\) by the chain rule gives
Consequently, at the point \((a,b)\), the slope of this branch is
Thus, when \(F_y(a,b)\ne 0\), the tangent is horizontal exactly when \(F_x(a,b)=0\). A zero slope gives the horizontal tangent line \(y=b\).
For a vertical tangent, use the other local graph description. If \(F_x(a,b)\ne 0\), the implicit function theorem gives a local branch \(x=x(y)\), and differentiating \(F(x(y),y)=0\) gives
If \(F_y(a,b)=0\), then \(dx/dy=0\): the curve is locally expressible as \(x=x(y)\) with zero slope in that description. Its tangent line is vertical, \(x=a\).
The condition that the other partial derivative is nonzero is essential. If both partial derivatives vanish, these tests do not classify the point. Such cases require separate analysis; vanishing one partial derivative alone is not enough to verify a tangent.
Why the Rule Works
The argument follows from differentiating the curve constraint along a local branch. The two possible graph descriptions give the two slope formulas. The following steps make explicit how the hypotheses connect to the conclusion.
Require \(F(a,b)=0\), so the point is part of the level curve being studied.
If \(F_y(a,b)\ne0\), write the curve locally as \(y(x)\). If \(F_x(a,b)\ne0\), write it locally as \(x(y)\).
The chain rule gives \(y'=-F_x/F_y\) or \(dx/dy=-F_y/F_x\), respectively.
When \(F_x=0\) and \(F_y\ne0\), \(y'=0\), so the tangent is horizontal. When \(F_y=0\) and \(F_x\ne0\), \(dx/dy=0\), so the tangent is vertical.
This also explains why the regularity check matters: the nonzero denominator is what permits the corresponding local graph and slope formula. The rule is not merely a pattern for guessing tangent directions; it is a consequence of the chain rule together with the local graph guaranteed by the implicit function theorem.
Worked Examples
Worked Example: Horizontal and Vertical Tangents to a Circle
Find and verify the horizontal and vertical tangents to \(F(x,y)=x^2+y^2-25=0\).
First compute the partial derivatives:
For horizontal candidates, solve \(F=0\) and \(F_x=0\). The condition \(2x=0\) gives \(x=0\). Substituting into the curve equation gives \(0^2+y^2-25=0\), so \(y^2=25\), and hence \(y=5\) or \(y=-5\). The candidates are \((0,5)\) and \((0,-5)\). At these points, \(F_y=10\) and \(F_y=-10\), respectively, both nonzero. Each point is on the curve, is regular, and has a horizontal tangent. The tangent lines are \(y=5\) and \(y=-5\).
For vertical candidates, solve \(F=0\) and \(F_y=0\). The condition \(2y=0\) gives \(y=0\). Then \(x^2-25=0\), so \(x=5\) or \(x=-5\). At \((5,0)\) and \((-5,0)\), \(F_x=10\) and \(F_x=-10\), respectively, both nonzero. These are verified regular points with vertical tangents \(x=5\) and \(x=-5\).
As a check, at \((0,5)\), \(y'=-F_x/F_y=-0/10=0\). At \((5,0)\), the other slope is \(dx/dy=-F_y/F_x=-0/10=0\). The points also satisfy the original equation: each has \(x^2+y^2=25\).
Worked Example: Tangents to a Tilted Ellipse
Classify the horizontal and vertical tangents to \(F(x,y)=x^2+xy+y^2-3=0\).
The partial derivatives are
For horizontal candidates, set \(F_x=0\), giving \(y=-2x\). Substitute this into the curve equation:
Thus \(x^2=1\), so \(x=1\) or \(x=-1\). The corresponding points are \((1,-2)\) and \((-1,2)\). Both lie on the curve: in each case \(x^2+xy+y^2=1-2+4=3\). At \((1,-2)\), \(F_y=1+2(-2)=-3\); at \((-1,2)\), \(F_y=-1+2(2)=3\). Neither value is zero, so both points are regular horizontal-tangent points. Their tangent lines are \(y=-2\) and \(y=2\).
For vertical candidates, set \(F_y=0\), giving \(x=-2y\). Substitution gives
Therefore \(y=1\) or \(y=-1\), and \(x=-2\) or \(x=2\), respectively. The points are \((-2,1)\) and \((2,-1)\). Each satisfies the curve equation because \(x^2+xy+y^2=4-2+1=3\). The other partial derivative is \(F_x=2x+y\), which equals \(-3\) at \((-2,1)\) and \(3\) at \((2,-1)\). These are nonzero, verifying regularity and vertical tangents \(x=-2\) and \(x=2\).
For a slope check, at \((1,-2)\), \(y'=-F_x/F_y=-0/(-3)=0\). At \((-2,1)\), \(dx/dy=-F_y/F_x=-0/(-3)=0\). These agree with the classifications.
Worked Example: A Vertical Tangent at the Origin
Determine whether \(F(x,y)=x-y^2=0\) has a vertical tangent at the origin.
Here \(F_x=1\) and \(F_y=-2y\). The origin lies on the curve because \(F(0,0)=0-0^2=0\). At the origin, \(F_y(0,0)=0\), whereas \(F_x(0,0)=1\ne0\). The vertical-tangent conditions are satisfied, and the point is regular.
We can also verify the branch directly. The equation is \(x=y^2\), so near the origin the curve is a graph \(x=x(y)=y^2\). Its derivative is \(dx/dy=2y\), which equals \(0\) at \(y=0\). Thus the tangent has constant \(x\)-coordinate \(x=0\), so it is vertical. The implicit formula gives the same result: \(dx/dy=-F_y/F_x=-0/1=0\) at the origin. Substitution into the original relation confirms that \(x=0,y=0\) is on the curve.
A Common Mistake: Reversing the Partial Derivatives
A frequent error is to associate \(F_x=0\) with a vertical tangent and \(F_y=0\) with a horizontal one. The partial derivatives are components of a normal direction to the curve, so it can be tempting to read their zeros as tangent directions directly. The slope formula shows why that assignment is reversed.
For example, take the circle \(x^2+y^2-4=0\) at \((0,2)\). Here \(F_x=2x=0\) and \(F_y=2y=4\). A mistaken classification might call this vertical because \(F_x=0\). But the valid local slope formula gives \(y'=-F_x/F_y=-0/4=0\), so the tangent is horizontal. The point lies on the circle since \(0^2+2^2-4=0\), and \(F_y=4\ne0\) verifies that the local \(y(x)\) branch exists. Its tangent line is \(y=2\).
A Reliable Procedure
When a problem asks for horizontal or vertical tangents of an implicit curve, organize the work so that candidate generation and verification remain distinct.
- Compute \(F_x\) and \(F_y\).
- For horizontal candidates, solve \(F(x,y)=0\) together with \(F_x(x,y)=0\).
- For every resulting point, evaluate \(F_y\). Keep it as a verified horizontal tangent only if \(F_y\ne0\).
- For vertical candidates, solve \(F(x,y)=0\) together with \(F_y(x,y)=0\).
- For every resulting point, evaluate \(F_x\). Keep it as a verified vertical tangent only if \(F_x\ne0\).
- State the tangent line: \(y=b\) for a horizontal tangent at \((a,b)\), or \(x=a\) for a vertical tangent.
Solving only \(F_x=0\) or \(F_y=0\) is not sufficient: the point must also lie on \(F=0\). Likewise, finding a point on the curve where one partial derivative is zero does not complete the argument until the other partial derivative is checked. That final check is what establishes a regular local branch and makes the tangent classification valid.
Why This Method Is Useful
Implicit equations often describe curves that are awkward or impossible to solve globally for \(y\) in terms of \(x\). Partial derivatives allow tangent directions to be found directly from the defining equation, without first rearranging it into an explicit formula. They also make the logical conditions visible: the curve equation locates the point, one vanishing partial identifies a candidate direction, and the other nonzero partial verifies the appropriate local graph.
The method therefore gives both an efficient calculation and a safeguard against overclaiming. A zero derivative condition produces candidates, not automatically valid tangents. Checking the point and regularity keeps the conclusion tied to the local geometry the implicit equation actually guarantees.
Check Your Understanding
For each question, state the conditions you would check before declaring a tangent direction.
- For a horizontal tangent to \(F(x,y)=0\) at \((a,b)\), which partial derivative must vanish, and which must be nonzero?
- Why must a tangent candidate satisfy \(F(a,b)=0\)?
- For a vertical tangent, which local graph description is used, and what derivative must vanish?
- At a point where both \(F_x\) and \(F_y\) vanish, can the classification rule in this tutorial verify a horizontal or vertical tangent? Explain.
- For \(F(x,y)=x-y^2\), verify the vertical tangent at the origin using \(F_x\), \(F_y\), and \(dx/dy\).