Tangent Lines to Implicit Curves
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.8: “Implicit Differentiation” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First
The curve $y^2 = x^3 - x + 1$ is called an elliptic curve. Plot or imagine its rough shape.
Predict: does the curve have a horizontal tangent anywhere? What would that require (in terms of $dy/dx$)?
Horizontal tangent requires $dy/dx = 0$. Implicit differentiation gives $2y\,dy/dx = 3x^2 - 1$, so $dy/dx = 0$ when $3x^2 - 1 = 0$, i.e., $x = \pm 1/\sqrt{3}$. Setting up and computing such slopes is what follows.
Prerequisite Check
Quick Reference
Finding the tangent line at $(x_0, y_0)$ on $F(x,y) = 0$:
- Verify $(x_0, y_0)$ is on the curve ($F(x_0, y_0) = 0$).
- Find $dy/dx$ implicitly (leave it as a formula in $x$ and $y$).
- Substitute $(x_0, y_0)$: $m = dy/dx\big|_{(x_0, y_0)}$.
- Tangent line: $y - y_0 = m(x - x_0)$.
Key Concepts
1. The Standard Procedure
Example 1. Find the tangent line to $x^2 + y^2 = 25$ at $(3, 4)$.
Step 1: verify $(3,4)$: $9 + 16 = 25$. On the curve.
Step 2: $2x + 2y\,dy/dx = 0$, so $dy/dx = -x/y$.
Step 3: at $(3,4)$: $m = -3/4$.
Step 4: $y - 4 = -\frac{3}{4}(x - 3)$, i.e., $y = -\frac{3}{4}x + \frac{25}{4}$.
Example 2. Find the tangent line to $x^3 + y^3 = 6xy$ at $(2, 2)$.
Step 1: $8 + 8 = 6 \cdot 2 \cdot 2 = 24$? No: $16 \neq 24$. But actually: $8 + 8 = 16$ and $6 \cdot 4 = 24$... that means $(2,2)$ is NOT on the curve.
Checking: $(3/2)^3 + (3/2)^3 = 27/4 + 27/4 = 27/2 \neq 3(3/2)(3/2) = 27/4$. Let me check $(3, 3/2)$: wait, instead, let’s use a point on the folium.
The folium $x^3 + y^3 = 3xy$ passes through $(3/2, 3/2)$: $(27/8 + 27/8) = 27/4$ and $3(3/2)(3/2) = 27/4$. Good.
Let me revise: find the tangent to $x^3 + y^3 = 3xy$ at $(3/2, 3/2)$.
Step 1: verified above.
Step 2: $3x^2 + 3y^2\,dy/dx = 3y + 3x\,dy/dx$. So $(3y^2 - 3x)\,dy/dx = 3y - 3x^2$, giving $dy/dx = (y - x^2)/(y^2 - x)$.
Step 3: at $(3/2, 3/2)$: $dy/dx = (3/2 - 9/4)/(9/4 - 3/2) = (-3/4)/(3/4) = -1$.
Step 4: $y - 3/2 = -1(x - 3/2)$, i.e., $y = -x + 3$.
2. Horizontal and Vertical Tangents
Horizontal tangent: $dy/dx = 0$. This requires the numerator of $dy/dx$ to be zero (with non-zero denominator).
Vertical tangent: $dy/dx$ is undefined (denominator is zero with non-zero numerator).
Example 3. Find all points on $x^2 + xy + y^2 = 7$ where the tangent is horizontal.
Differentiate: $2x + y + x\,dy/dx + 2y\,dy/dx = 0$.
$(x + 2y)\,dy/dx = -(2x + y)$.
$dy/dx = 0$ when $2x + y = 0$, i.e., $y = -2x$.
Substitute $y = -2x$ into $x^2 + xy + y^2 = 7$:
$x^2 + x(-2x) + (-2x)^2 = 7$: $x^2 - 2x^2 + 4x^2 = 3x^2 = 7$, so $x = \pm\sqrt{7/3}$.
Horizontal tangents at $\left(\sqrt{7/3}, -2\sqrt{7/3}\right)$ and $\left(-\sqrt{7/3}, 2\sqrt{7/3}\right)$.
3. Multiple Representations: Graph, Formula, and Table
For $x^2 + y^2 = 25$:
| Point | $dy/dx = -x/y$ | Geometric meaning |
|---|---|---|
| $(0, 5)$ | $0$ | Horizontal tangent at top of circle |
| $(3, 4)$ | $-3/4$ | Declining to the right |
| $(4, 3)$ | $-4/3$ | More steeply declining |
| $(5, 0)$ | undefined | Vertical tangent at rightmost point |
| $(-3, 4)$ | $3/4$ | Rising to the right |
The formula $-x/y$ predicts all of these simultaneously. Explicit solving would require a separate calculation for each branch.
4. Normal Lines
The normal line at $(x_0, y_0)$ is perpendicular to the tangent. Its slope is $-1/m$ (where $m = dy/dx$ at the point).
Example 4. Normal to $x^2 + y^2 = 25$ at $(3, 4)$:
Tangent slope $= -3/4$. Normal slope $= 4/3$.
Normal line: $y - 4 = \frac{4}{3}(x - 3)$, i.e., $y = \frac{4}{3}x$. (Passes through the origin, as expected for a circle’s normal line.)
5. Ask Why: Why Does the Normal to a Circle Pass Through the Center?
The center of $x^2 + y^2 = r^2$ is the origin. The radius to $(x_0, y_0)$ has slope $y_0/x_0$. The tangent has slope $-x_0/y_0$ (implicit differentiation). The normal has slope $y_0/x_0$ (negative reciprocal of tangent slope). The normal at $(x_0, y_0)$ with slope $y_0/x_0$ passes through the origin: $y - y_0 = (y_0/x_0)(x - x_0)$ gives $y = (y_0/x_0)x$ at $x = 0, y = 0$. Confirmed.
The normal is always a radius. This is a deep geometric fact: perpendicularity to the tangent is the geometric definition of a radius.
Named Misconception: Not Verifying the Point Is on the Curve
A common error is computing the tangent slope without first checking that the given point satisfies the equation. If the point is not on the curve, the slope formula gives a number but the “tangent line” is not tangent to anything.
Always verify: substitute $(x_0, y_0)$ into $F(x, y) = c$ before proceeding.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Not verifying the point | Computing slope at a point not on the curve | Check $F(x_0, y_0) = c$ first |
| Evaluating $dy/dx$ before isolating it | Substituting before solving for $dy/dx$ | Find the general formula first, then substitute |
| Normal slope error | Using $m$ instead of $-1/m$ | Normal slope $= -1/(dy/dx)$ |
Common Misconceptions
the value of $dy/dx$ at a point equals the $y$-coordinate of the point.
This is the height-vs-slope error. For the circle $x^2 + y^2 = 25$ at the point $(3, 4)$, the height is $f(3) = 4$ and the slope is $dy/dx = -3/4$. These are different numbers with different meanings: $4$ is a position on the curve, while $-3/4$ is the steepness of the tangent line. Confusing them leads to tangent line equations with incorrect slopes, because the formula $dy/dx = -x/y$ must be evaluated at the point to yield the slope, not read off as the $y$-value.
Leveled Practice
Level 1 -- Tangent Line at a Given Point
Problem 1. Find the tangent line to $x^2 + 4y^2 = 8$ at $(2, 1)$.
Show answer
Verify: $4 + 4 = 8$. On curve.
Differentiate: $2x + 8y\,dy/dx = 0 \Rightarrow dy/dx = -x/(4y)$.
At $(2,1)$: $m = -2/4 = -1/2$.
Tangent: $y - 1 = -\frac{1}{2}(x-2)$, i.e., $y = -\frac{x}{2} + 2$.
Problem 2. Find the tangent line to $y^3 + xy = 5$ at $(4, 1)$.
Show answer
Verify: $1 + 4 = 5$. On curve.
Differentiate: $3y^2\,dy/dx + y + x\,dy/dx = 0 \Rightarrow (3y^2 + x)\,dy/dx = -y \Rightarrow dy/dx = -y/(3y^2+x)$.
At $(4,1)$: $dy/dx = -1/(3+4) = -1/7$.
Tangent: $y - 1 = -\frac{1}{7}(x-4)$.
Level 2 -- Horizontal and Vertical Tangents
Problem 3. Find all points on $x^2 - xy + y^2 = 3$ where the tangent is horizontal.
Show answer
Differentiate: $2x - y - x\,dy/dx + 2y\,dy/dx = 0 \Rightarrow (2y-x)\,dy/dx = y - 2x \Rightarrow dy/dx = (y-2x)/(2y-x)$.
Horizontal: $y - 2x = 0 \Rightarrow y = 2x$.
Substitute: $x^2 - 2x^2 + 4x^2 = 3x^2 = 3 \Rightarrow x = \pm 1$.
Points: $(1, 2)$ and $(-1, -2)$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Find the tangent to the unit circle $x^2 + y^2 = 1$ at $(1/\sqrt{2}, 1/\sqrt{2})$ and verify it is perpendicular to the radius.
(b) (Mid) Show that the tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ at $(x_0, y_0)$ has equation $\dfrac{x_0 x}{a^2} + \dfrac{y_0 y}{b^2} = 1$.
(c) (Ceiling) For the curve $x^2 y + y^3 = 2$, find all points where the tangent line is vertical.
Show answer
(a) $dy/dx = -x/y = -1$ at the point. Radius slope $= 1$. Product $(-1)(1) = -1$. Perpendicular. Tangent: $y - 1/\sqrt{2} = -(x - 1/\sqrt{2})$, i.e., $x + y = \sqrt{2}$.
(b) Implicit: $\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$. At $(x_0, y_0)$: slope $= -b^2 x_0/(a^2 y_0)$.
Tangent: $y - y_0 = -\frac{b^2 x_0}{a^2 y_0}(x - x_0)$.
Rearrange: $a^2 y_0(y - y_0) = -b^2 x_0(x - x_0)$, i.e., $a^2 y_0 y + b^2 x_0 x = a^2 y_0^2 + b^2 x_0^2 = a^2 b^2$ (since $(x_0,y_0)$ is on the ellipse). Dividing by $a^2 b^2$: $\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1$.
(c) Differentiate $x^2 y + y^3 = 2$: $2xy + x^2 y' + 3y^2 y' = 0 \Rightarrow y'(x^2 + 3y^2) = -2xy \Rightarrow y' = -2xy/(x^2+3y^2)$.
Vertical tangent: $x^2 + 3y^2 = 0$. For real solutions: $x = y = 0$. But $(0,0)$: $0 + 0 = 2$, not on curve. No vertical tangents.
Mastery Checklist
Mental Model
The tangent line to an implicit curve is found the same way as for an explicit function -- compute $dy/dx$ at the point, use point-slope form -- but the computation of $dy/dx$ uses implicit differentiation.
The key new twist: the slope formula contains both $x$ and $y$. At different points on the same curve, you get different slopes from the same formula, which is why one compact formula $dy/dx = -x/y$ covers all points on the circle simultaneously.
Connections
Within MATH161
- Second derivatives implicitly: Differentiating $dy/dx$ a second time gives $d^2y/dx^2$, needed for concavity.
- Related rates: The same setup -- differentiating an implicit equation -- applies to related-rate problems where $x$ and $y$ both depend on time.
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