Tangent lines to curves defined implicitly

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is the circle $x^2 + y^2 = 25$ ; the blue line is its tangent at the point at angle $a$ . Move $a$ .

The circle x² + y² = 25 and its tangent line at the point (5 cos a, 5 sin a). The tangent is perpendicular to the radius, with slope −x/y.

The radius to $(3, 4)$ has slope $\tfrac{4}{3}$ . What is the slope of the tangent there?

$-\tfrac{3}{4}$
$\tfrac{4}{3}$

Key idea

After implicit differentiation, $y'$ depends on both coordinates.

To find the tangent at

(x_0, y_0)

: compute

y'

implicitly, substitute both

x_0

and

y_0

to get the slope

m

, then

y = y_0 + m(x - x_0)

.

Worked example

Tangent to $x^2 + xy + y^2 = 7$ at $(1, 2)$ .

2x + y + x\,y' + 2y\,y' = 0

Product rule on xy.

y' = -\dfrac{2x + y}{x + 2y}

Solve for y'.

y = 2 - \tfrac{4}{5}(x - 1)

At (1, 2): −4/5.

Your turn, guided

Your turn: $x^2 + y^2 = 13$ at $(2, 3)$ .

slope

= -\tfrac{2}{3}

y' = −x/y at (2, 3).

y = 3 - \tfrac{2}{3}(x - 2)

Point-slope form.

Practice ★

Find the slope of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(4, 3)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 40$ at the point $(2, 6)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 26$ at the point $(1, 5)$ .

Watch out!

Practice ★★

Write the tangent line to $x^2 + y^2 = 20$ at $(4, 2)$ : enter the right-hand side of $y = \ldots$
Write the tangent line to $x^2 + y^2 = 45$ at $(6, 3)$ : enter the right-hand side of $y = \ldots$

The point $(1, 2)$ is on the curve $x^2 + xy + y^2 = 7$ . Find the slope of the tangent line there.
The point $(1, 5)$ is on the curve $x^2 + xy + y^2 = 31$ . Find the slope of the tangent line there.

Exit ticket

Three questions to check you have got it.

Find the slope of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(4, 3)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 40$ at the point $(2, 6)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 26$ at the point $(1, 5)$ .

Summary card

Tangents to implicit curves

Slope: substitute the point into $y'$ (both $x$ and $y$ ).
Line: $y = y_0 + m(x - x_0)$ .

Watch out: the tangent to a circle is perpendicular to the radius.

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