Find horizontal and vertical tangents
Where the tangent is flat or upright
The curve is \(x^2 - xy + y^2 = 12\), and its slope is the fraction \(\tfrac{dy}{dx} = \tfrac{y - 2x}{2y - x}\).
Go to the top at \((2, 4)\). The numerator is \(4 - 4 = 0\) and the denominator is \(8 - 2 = 6\), so the slope is 0 and the tangent is horizontal.
Go to the right at \((4, 2)\). The denominator is \(4 - 4 = 0\) and the numerator is \(2 - 8 = -6\), so the slope does not exist and the tangent is vertical.
A fraction is 0 when its numerator is 0, and it has no value when its denominator is 0.
Any fraction
Remove the numbers. The slope is a fraction \(\tfrac{dy}{dx} = \tfrac{N}{D}\), where \(N\) and \(D\) depend on \(x\) and \(y\).
Drag the point and the tangent turns. It is flat where \(N = 0\) and \(D\) is not 0, because the slope is 0 there.
It is upright where \(D = 0\) and \(N\) is not 0. The slope does not exist there, but the tangent is the vertical line through the point.
These conditions only tell you where to look. The point must also be on the curve.
Notation
Write the slope as a fraction. For the curve \(y^2 = x^2 + x^3\) in the figure, \(N = 2x + 3x^2\) and \(D = 2y\).
\[\frac{dy}{dx} = \frac{\cy{N}}{\cy{D}}\]For a horizontal tangent, solve \(N = 0\). Then check that \(D\) is not 0 at each point.
\[N = 0, \qquad D \neq 0\]For a vertical tangent, solve \(D = 0\). Then check that \(N\) is not 0.
\[D = 0, \qquad N \neq 0\]Each condition gives a relation between \(x\) and \(y\), not a point. Here \(N = x(2 + 3x)\) is 0 at \(x = 0\), the origin, and at \(x = -\tfrac23\), where the dashed line meets the curve twice. Put \(x = -\tfrac23\) in the equation of the curve to find these two points.
\[x = -\tfrac23 \quad\Rightarrow\quad y^2 = \tfrac{4}{27}\]If no point of the curve satisfies the relation, there is no such tangent.
When it fails
For \(y^2 = x^2 + x^3\), the slope is \(\tfrac{2x + 3x^2}{2y}\). At the origin, \(N\) and \(D\) are both 0, so the fraction is \(\tfrac00\) and gives no slope.
The curve crosses itself there, with two tangent lines of slopes 1 and \(-1\), shown in red. Neither is horizontal or vertical, so when \(N\) and \(D\) are both 0, neither condition applies.
Example 1: Horizontal and vertical
Find the points of \(x^2 + 2y^2 - 4x = 12\) where the tangent is horizontal or vertical.
Differentiate and solve for the slope.
\[\begin{aligned} 2x + 4y\,\frac{dy}{dx} - 4 &= 0 \\ \frac{dy}{dx} &= \frac{2 - x}{2y} \end{aligned}\]For a horizontal tangent, set the numerator to 0, which gives \(x = 2\). Put \(x = 2\) in the curve to find the two points where the dashed line meets it. The denominator \(2y\) is not 0 at either point.
\[\begin{aligned} 4 + 2y^2 - 8 &= 12 \\ y^2 &= 8 \\ (2, \cy{2\sqrt2}), \; &(2, \cy{-2\sqrt2}) \end{aligned}\]For a vertical tangent, set the denominator to 0, which gives \(y = 0\). Put \(y = 0\) in the curve. The numerator \(2 - x\) is \(-4\) or 4 there, so it is not 0.
\[\begin{aligned} x^2 - 4x - 12 &= 0 \\ x &= 6 \text{ or } x = -2 \end{aligned}\]
Example 2: A condition with no point
Find the points of \(x^2 + 2x - y^2 = 3\) where the tangent is horizontal or vertical.
Differentiate and solve for the slope.
\[\begin{aligned} 2x + 2 - 2y\,\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= \frac{x + 1}{y} \end{aligned}\]A tempting wrong move: For a horizontal tangent, set the numerator to 0, which gives \(x = -1\). The tangent is horizontal at \(x = -1\).
\[x = -1\]The dashed vertical line at \(x = -1\) does not meet the curve. Putting \(x = -1\) in the curve gives \(1 - 2 - y^2 = 3\), so \(y^2 = -4\), which no number satisfies. The condition only tells you where to look, and the point must also be on the curve.
The curve has no point with \(x = -1\), so it has no horizontal tangent.
For a vertical tangent, set the denominator to 0, which gives \(y = 0\). Put \(y = 0\) in the curve. The numerator \(x + 1\) is 2 or \(-2\) there, so it is not 0.
\[\begin{aligned} x^2 + 2x - 3 &= 0 \\ x &= 1 \text{ or } x = -3 \end{aligned}\]
Practice
Find the points of \(y^2 = x^3 - 3x\) where the tangent is horizontal.
Set the numerator to 0 and solve for \(x\). Then use the dashed vertical line to see whether the curve has points at that \(x\).
Solution
Differentiate and solve for the slope.
\[\begin{aligned} 2y\,\frac{dy}{dx} &= 3x^2 - 3 \\ \frac{dy}{dx} &= \frac{3x^2 - 3}{2y} \end{aligned}\]The numerator is 0 when \(x = 1\) or \(x = -1\).
At \(x = 1\), the curve gives \(y^2 = -2\), which has no solution. At \(x = -1\), it gives \(y^2 = 2\), so \(y = \pm\sqrt2\).
\[(-1, \sqrt2), \qquad (-1, -\sqrt2)\]The denominator \(2y\) is not 0 at these points, so the tangent is horizontal at both.
Questions
The figure shows the ellipse \(4x^2 + y^2 - 8x = 0\), whose slope is \(\tfrac{dy}{dx} = \tfrac{4(1 - x)}{y}\). Four points have bars that show a flat or an upright tangent.
Which pair of points has vertical tangents?
- \((0, 0)\) and \((2, 0)\)
- \((1, 2)\) and \((1, -2)\)
- \((0, 0)\) and \((1, 2)\)
Answer
\((0, 0)\) and \((2, 0)\)
What is the equation of the tangent line at \((0, 0)\)?
- \(x = 0\)
- \(y = 0\)
- There is no tangent, because the slope does not exist.
Answer
\(x = 0\)
The circle \((x - 2)^2 + (y - 5)^2 = 9\) has slope \(\tfrac{dy}{dx} = -\tfrac{x - 2}{y - 5}\). At which points is the tangent vertical?
- \((5, 5)\) and \((-1, 5)\)
- \((2, 8)\) and \((2, 2)\)
- \((2, 5)\)
Answer
\((5, 5)\) and \((-1, 5)\)
Summary
You set the numerator of the slope to 0 for horizontal tangents, and the denominator to 0 for vertical tangents.
This works because a fraction is 0 when its numerator is 0, and it has no value when its denominator is 0.
On an implicit curve, a solution counts only if it lies on the curve and the other part of the fraction is not 0.
This is one of four skills in topic 3.2 of Which & Why: AP Calculus.