Single-Variable-Calculus · Unit 1 · Video 4 · Interactive Practice

Tangents to 1/x and the Triangle of Area 2

IKey Formulas

Formula	Name	What you need
$y - y_0 = -\dfrac{1}{x_0^2}\,(x - x_0)$	Tangent line to $y = 1/x$ — the starred equation $(*)$	Point-slope form and the slope $f'(x_0) = -1/x_0^2$
$x\text{-intercept} = 2x_0$	Base of the triangle	Set $y = 0$ in $(*)$ , using $y_0 = 1/x_0$
$y\text{-intercept} = 2y_0 = \dfrac{2}{x_0}$	Height of the triangle	Set $x = 0$ in $(*)$ , or swap $x \leftrightarrow y$ in the base result
$A = \tfrac{1}{2}(2x_0)(2y_0) = 2x_0y_0 = 2$	Area of the tangent triangle	Both intercepts and $x_0y_0 = 1$

Key Insight: Exactly one line of this problem is calculus — the slope $-1/x_0^2$ . Point-slope form, clearing fractions, the symmetry of $xy = 1$ and $\tfrac{1}{2}bh$ are all older than the course, and the free parameter $x_0$ cancels out of the final answer.

IIThe Triangle That Never Changes

The point of tangency is a free choice, yet every tangent to $y = 1/x$ cuts off the same area.

Tangency point x₀ 1.25

💡 The same computation on $y = c/x$ gives every one of its tangent triangles the area $2c$ ; this curve is the case $c = 1$ .

IIIWhere the Two Intercepts Come From

One calculus step supplies the slope; the intercepts $2x_0$ and $2y_0$ then fall out of algebra alone.

Step 1

Step 1 — The only calculus: the slope

x_0 = 2

the point of tangency is

(x_0, y_0) = \left(2, \tfrac{1}{2}\right)

, and

f'(x_0) = -\dfrac{1}{x_0^2} = -\dfrac{1}{4}

(*)\qquad y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2)

Step 2 — The base: set $y = 0$

The

x

-axis is the line

y = 0

, and the tangency point sits on the curve, so

y_0 = 1/x_0

0 - \frac{1}{x_0} = -\frac{1}{x_0^2}\,(x - x_0)

Step 3 — Clear the fractions

-\frac{1}{x_0} = -\frac{1}{x_0^2}\,x + \frac{1}{x_0} \quad\Longrightarrow\quad \frac{1}{x_0^2}\,x = \frac{2}{x_0}

x = x_0^2\cdot\frac{2}{x_0} = 2x_0 = 4

Step 4 — The height, for free

y = 1/x

and

x = 1/y

are the same equation

xy = 1

, so the picture is its own mirror image across

y = x

: every true formula survives the swap

x \leftrightarrow y

x = 2x_0 \quad\longrightarrow\quad y = 2y_0 = \frac{2}{x_0} = 1

Step 5 — Half base times height

A = \tfrac{1}{2}(2x_0)(2y_0) = 2x_0y_0 = 2x_0\cdot\frac{1}{x_0} = 2

Here

\tfrac{1}{2}(4)(1) = 2

— and no

x_0

survives to the answer.

IVOne Letter, Two Objects

The same letter $y$ names the hyperbola in one equation and the horizontal axis two lines later.

VQuiz Questions

Problem 1 · Reading Off the Intercepts

Given: the tangent to $y = \dfrac{1}{x}$ at the point where $x_0 = 4$ — find where that tangent line crosses the two axes.

✅ Correct! Each intercept doubles a coordinate of the tangency point:

2x_0 = 8

and

2y_0 = \tfrac{1}{2}

, so the area is

\tfrac{1}{2}(8)\left(\tfrac{1}{2}\right) = 2

❌ Close, but… those are the tangency point's own coordinates dropped onto the axes. Solving

(*)

with

y = 0

pushes the crossing out to

x = 2x_0

, twice as far.

❌ Not quite. The base lies on the

x

-axis and must be

2x_0 = 8

; you have the two legs exchanged. That swap is the tangent triangle at

x_0 = \tfrac{1}{4}

, not at

x_0 = 4

❌ Not quite. Substitute

y = 0

into

y - \tfrac{1}{4} = -\tfrac{1}{16}(x - 4)

and solve for

x

; the slope

-1/x_0^2

is used once, not squared again.

Show solution

At $x_0 = 4$ : $y_0 = \tfrac{1}{4}$ and $f'(4) = -\dfrac{1}{4^2} = -\dfrac{1}{16}$ , so

(*)\qquad y - \tfrac{1}{4} = -\tfrac{1}{16}(x - 4)

$x$ -intercept — set $y = 0$ :

-\tfrac{1}{4} = -\tfrac{1}{16}x + \tfrac{1}{4} \quad\Longrightarrow\quad \tfrac{1}{16}x = \tfrac{1}{2} \quad\Longrightarrow\quad x = 8 = 2x_0

$y$ -intercept — set $x = 0$ :

y = \tfrac{1}{4} + \tfrac{1}{16}(4) = \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2} = 2y_0

Area $= \tfrac{1}{2}(8)\left(\tfrac{1}{2}\right) = 2$ , as it must be.

Problem 2 · The Slope Is Where the Signs Hide

Given: the curve $y = \dfrac{1}{x}$ at $x_0 = \tfrac{1}{2}$ — find the point-slope equation of the tangent line there.

✅ Correct!

y_0 = 1/x_0 = 2

and

f'\left(\tfrac{1}{2}\right) = -\dfrac{1}{(1/2)^2} = -4

. A small

x_0

makes the tangent very steep — the tall, thin triangle.

❌ Close, but… you used

-1/x_0

instead of

-1/x_0^2

. Squaring first gives

(1/2)^2 = 1/4

, and

-1/(1/4) = -4

❌ Check the sign. The hyperbola falls throughout the first quadrant, so every tangent slope there is negative:

-1/x_0^2 < 0

❌ Not quite. The coordinates are exchanged: the point of tangency is

(x_0, y_0) = \left(\tfrac{1}{2}, 2\right)

, so

x_0 = \tfrac{1}{2}

belongs in the

x

-slot.

❌ Not quite. Build the line from

(x_0, y_0) = \left(\tfrac{1}{2}, 2\right)

and the slope

-1/x_0^2

Show solution

Step 1 — the point. The tangency point lies on the curve, so $y_0 = \dfrac{1}{x_0} = \dfrac{1}{1/2} = 2$ , giving $\left(\tfrac{1}{2},\, 2\right)$ .

Step 2 — the slope (the whole calculus content):

f'\left(\tfrac{1}{2}\right) = -\frac{1}{(1/2)^2} = -\frac{1}{1/4} = -4

Step 3 — point-slope form:

y - 2 = -4\left(x - \tfrac{1}{2}\right)

Check the intercepts: setting $y = 0$ gives $x = 1 = 2x_0$ ; setting $x = 0$ gives $y = 4 = 2y_0$ . Area $= \tfrac{1}{2}(1)(4) = 2$ . ✅

Problem 3 · Running the Argument Backwards

Given: a tangent line to the first-quadrant branch of $y = \dfrac{1}{x}$ crosses the $x$ -axis at $x = 6$ — find its point of tangency and the height of the triangle it cuts off.

Where does the line touch the curve?

How tall is the triangle?

✅ Correct!

2x_0 = 6

forces

x_0 = 3

, hence

y_0 = \tfrac{1}{3}

and a height of

2y_0 = \tfrac{2}{3}

— and

\tfrac{1}{2}(6)\left(\tfrac{2}{3}\right) = 2

, as promised.

❌ Check the tangency point. The intercept is

2x_0

, not

x_0

: an

x

-intercept of

6

means

x_0 = 3

, and then

y_0 = 1/x_0

❌ Check the height. The height is the

y

-intercept

2y_0

— double the tangency height

y_0

, not

y_0

itself.

Show solution

Step 1 — invert the base formula. Every such tangent has $x$ -intercept $2x_0$ , so

2x_0 = 6 \quad\Longrightarrow\quad x_0 = 3

Step 2 — land the point on the curve: $y_0 = \dfrac{1}{x_0} = \dfrac{1}{3}$ , so the tangency point is $\left(3, \tfrac{1}{3}\right)$ .

Step 3 — the height is the $y$ -intercept:

2y_0 = \frac{2}{x_0} = \frac{2}{3}

Check with the tangent line. The slope is $-\dfrac{1}{3^2} = -\dfrac{1}{9}$ , so $y - \tfrac{1}{3} = -\tfrac{1}{9}(x - 3)$ . Setting $x = 0$ gives $y = \tfrac{1}{3} + \tfrac{1}{3} = \tfrac{2}{3}$ ✓, and setting $y = 0$ gives $x = 6$ ✓.

Area $= \tfrac{1}{2}(6)\left(\tfrac{2}{3}\right) = 2$ .

Problem 4 · The Whole Family $y = c/x$

Given: a curve $y = \dfrac{c}{x}$ with $c > 0$ , whose tangent lines all cut off triangles of area $10$ — find $c$ , then locate one intercept.

What is $c$ ?

For that curve, where does the tangent at $x_0 = 3$ cross the $x$ -axis?

✅ Correct! Area

= 2c

gives

c = 5

, while the base

2x_0 = 6

knows nothing about

c

: stretching the curve vertically moves the height, never the

x

-intercept.

❌ Check the constant. Repeating the derivation with

y = c/x

gives base

2x_0

and height

2c/x_0

, so the area is

\tfrac{1}{2}(2x_0)\left(\tfrac{2c}{x_0}\right) = 2c

— set that equal to

10

❌ Check the base. The

c

cancels out of the base computation entirely: the

x

-intercept is

2x_0

for every value of

c

Show solution

Step 1 — redo the derivation with the constant carried along. Running the same difference quotient with $c$ carried through gives $f'(x_0) = -\dfrac{c}{x_0^2}$ , and $y_0 = \dfrac{c}{x_0}$ , so

y - \frac{c}{x_0} = -\frac{c}{x_0^2}(x - x_0)

Step 2 — the intercepts. Setting $y = 0$ :

-\frac{c}{x_0} = -\frac{c}{x_0^2}x + \frac{c}{x_0} \quad\Longrightarrow\quad \frac{c}{x_0^2}x = \frac{2c}{x_0} \quad\Longrightarrow\quad x = 2x_0

The $c$ cancels — the base is $2x_0$ no matter what $c$ is. Setting $x = 0$ gives $y = \dfrac{2c}{x_0} = 2y_0$ .

Step 3 — the area:

A = \tfrac{1}{2}(2x_0)\left(\frac{2c}{x_0}\right) = 2c

Step 4 — answer both parts. $2c = 10$ gives $c = 5$ , and at $x_0 = 3$ the $x$ -intercept is $2x_0 = 6$ (the height there is $\tfrac{2(5)}{3} = \tfrac{10}{3}$ , and $\tfrac{1}{2}(6)\left(\tfrac{10}{3}\right) = 10$ ✓).

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