Why Look for Vertical Tangents?
A tangent line describes the direction in which a curve is moving at a point. For a curve written as \(y\) in terms of \(x\), a vertical tangent is a place where the curve’s direction is vertical rather than having a finite slope \(dy/dx\). An equation may define a curve without giving \(y\) explicitly as a function of \(x\), so implicit differentiation provides a useful way to locate such points.
The key is to inspect the implicit slope formula carefully. A zero denominator alone is not enough: the point must lie on the curve, and the numerator must be nonzero. When those conditions hold for a smooth relation, the curve has a vertical tangent there.
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. If \(y\) is locally a differentiable function of \(x\), differentiating \(F(x,y)=0\) with respect to \(x\) gives
Here \(F_x\) and \(F_y\) denote the partial derivatives of \(F\) with respect to \(x\) and \(y\). When \(F_y(x,y)\ne 0\), solving for the slope gives
In this quotient, the numerator is \(-F_x\) and the denominator is \(F_y\). Thus the standard conditions for a candidate vertical tangent at \((a,b)\) are
The first condition places the point on the curve. The second makes the denominator of the slope formula zero, and the third ensures the numerator is nonzero. Together, the last two conditions mean that the implicit slope quotient is not a finite number there. To interpret the geometry, we also need to know why they indicate a vertical tangent, rather than relying only on the quotient.
Why the Conditions Give a Vertical Tangent
When \(F_x(a,b)\ne 0\), the implicit function theorem says that the curve can locally be written as \(x=g(y)\) near \((a,b)\). In other words, close to the point, \(x\) is a differentiable function of \(y\). Differentiating \(F(g(y),y)=0\) with respect to \(y\) gives
Because \(F_x(a,b)\ne 0\), this equation gives the local derivative at the point:
The curve can therefore be represented locally by the parametrization \((g(y),y)\). Its tangent direction at \(y=b\) is \((g'(b),1)=(0,1)\), which is vertical. This justifies the criterion: the curve has the vertical tangent line \(x=a\). The condition \(F_x(a,b)\ne 0\) matters because it permits this local representation and gives a nonzero vertical component in the tangent direction.
A Reliable Search Procedure
Find an expression for \(dy/dx\) and identify its numerator and denominator.
Solve for possible coordinates, keeping any domain restrictions in view.
Discard any resulting coordinate pair that does not satisfy \(F(x,y)=0\).
For each point on the curve, confirm that the numerator is nonzero.
Under the stated smoothness conditions, give the vertical tangent line \(x=a\) at each verified point \((a,b)\).
Worked Examples
Worked Example: The Unit Circle
Find the vertical tangents to \(x^2+y^2=1\).
Differentiate both sides with respect to \(x\). The derivative of \(x^2\) is \(2x\), and the chain rule gives \(2y\,dy/dx\) for \(y^2\):
Solving for the slope gives
The denominator is zero when \(y=0\). Substituting \(y=0\) into the original equation gives \(x^2=1\), so \(x=1\) or \(x=-1\). The candidate points are \((1,0)\) and \((-1,0)\), and both satisfy the circle equation.
At \((1,0)\), the numerator \(-x=-1\) is nonzero. At \((-1,0)\), the numerator \(-x=1\) is also nonzero. The criterion is met at both points, so the vertical tangent lines are \(x=1\) and \(x=-1\).
For a geometric check, near either point the circle can be written as \(x=\pm\sqrt{1-y^2}\), using the branch matching that point. Differentiating gives \(dx/dy=\mp y/\sqrt{1-y^2}\), which equals \(0\) at \(y=0\). The tangent direction is therefore vertical at each point, as predicted.
Worked Example: A Curve with a Cubic Term
Find the vertical tangents to \(x^2+y^3=4\).
Implicit differentiation gives
The denominator \(3y^2\) is zero when \(y=0\). Substitution into the original equation gives \(x^2=4\), so \(x=2\) or \(x=-2\). The candidate points \((2,0)\) and \((-2,0)\) both lie on the curve. Their numerators are \(-2x=-4\) and \(-2x=4\), respectively, so neither numerator is zero.
Thus both points have vertical tangents, with equations \(x=2\) and \(x=-2\). To check the local geometry, solve the original relation for \(x\) near each point:
For the positive branch, \(dx/dy=-3y^2/(2\sqrt{4-y^3})\), which equals \(0\) at \(y=0\). For the negative branch, \(dx/dy=3y^2/(2\sqrt{4-y^3})\), which also equals \(0\) at \(y=0\). Each parametrization therefore has tangent direction \((0,1)\) at its point.
Worked Example: Finding More Than One Candidate
Find the vertical tangents to \(x^2+xy+y^2=3\).
Differentiate each term with respect to \(x\). In particular, the product rule gives \(d(xy)/dx=y+x\,dy/dx\):
Collect the derivative terms and solve:
The denominator is zero when \(x+2y=0\), so \(x=-2y\). Substitute this into the original equation:
Thus \(y^2=1\), so \(y=1\) or \(y=-1\). The corresponding \(x\)-coordinates are \(x=-2\) and \(x=2\). Both points satisfy the original relation: at \((-2,1)\), \(4-2+1=3\); at \((2,-1)\), \(4-2+1=3\).
The numerator \(-(2x+y)\) is \(3\) at \((-2,1)\), since \(-(2(-2)+1)=3\). At \((2,-1)\), it is \(-3\), since \(-(2(2)-1)=-3\). Both are nonzero. Therefore the vertical tangents are \(x=-2\) at \((-2,1)\), and \(x=2\) at \((2,-1)\).
As a check, the curve can be solved locally for \(x\) as a function of \(y\):
At \((-2,1)\), use the minus sign. Differentiating that branch gives \(dx/dy=(-1+3y/\sqrt{12-3y^2})/2\), which at \(y=1\) is \((-1+3/3)/2=0\). At \((2,-1)\), use the plus sign. Its derivative is \((-1-3y/\sqrt{12-3y^2})/2\), which at \(y=-1\) is \((-1+3/3)/2=0\). These checks confirm the vertical tangent directions.
A Common Mistake: Calling a Zero Denominator a Zero Slope
For the unit circle, the slope formula is \(dy/dx=-x/y\). At \((1,0)\), a tempting but incorrect step is to see \(y=0\) in the denominator and declare the slope to be \(0\). A denominator of zero does not make a quotient zero; here the quotient is undefined because its numerator is \(-1\), not zero.
The correction is to check both parts of the quotient. At \((1,0)\), the point lies on the circle, the denominator \(y\) is zero, and the numerator \(-x=-1\) is nonzero. The criterion therefore identifies a vertical tangent, namely \(x=1\). A zero numerator with a nonzero denominator would instead give a horizontal tangent.
Why the Geometric Check Matters
The quotient formula is a convenient way to search, but its geometric interpretation depends on the conditions behind it. Checking the original equation ensures that a coordinate obtained from the denominator actually belongs to the curve. Checking the numerator ensures that the slope equation has not reduced to a different indeterminate case. Finally, \(F_x\ne 0\) provides the local representation \(x=g(y)\) that confirms the direction is vertical.
This process is useful whenever a curve is given implicitly and cannot conveniently be treated as a single explicit function \(y=f(x)\). It also encourages a precise distinction between a point where the formula is undefined and a point where the curve’s geometry has been established.
Check Your Understanding
For each question, check that any proposed point lies on the curve before interpreting the slope formula.
- For \(x^2+4y^2=16\), find the points where the implicit slope denominator is zero. What are the vertical tangent lines?
- For \(x^2+y^3=9\), find the candidate vertical tangent points and verify that the numerator is nonzero at each.
- For \(x^2+xy+y^2=3\), what equation must the coordinates satisfy for the slope denominator to be zero?
- Why is a nonzero numerator needed in the vertical-tangent criterion?
- What additional analysis is needed if both the numerator and denominator of the implicit slope formula are zero?
