### Equation of a Tangent Line in Cartesian Coordinates

Suppose that a function \(y = f\left( x \right)\) is defined on the interval \(\left( {a,b} \right)\) and is continuous at \({x_0} \in \left( {a,b} \right).\) At this point (the point \(M\) in Figure \(1\)), the function has the value \({y_0} = f\left( {{x_0}} \right).\)

Let the independent variable at \({x_0}\) has the increment \(\Delta x.\) The corresponding increment of the function \(\Delta y\) is expressed as

\[\Delta y = f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right).\]

In Figure \(1,\) the point \({M_1}\) has the coordinates \(\left( {{x_0} + \Delta x,{y_0} + \Delta y} \right).\) We draw the secant \(M{M_1}.\) Its equation has the form

\[y – {y_0} = k\left( {x – {x_0}} \right),\]

where \(k\) is the slope coefficient depending on the increment \(\Delta x\) and equal

\[k = k\left( {\Delta x} \right) = \frac{{\Delta y}}{{\Delta x}}.\]

When \(\Delta x\) decreases, the point \({M_1}\) moves to the point \(M:\) \({M_1} \to M.\) In the limit \(\Delta x \to 0\) the distance between the points \(M\) and \({M_1}\) approaches zero. This follows from the continuity of the function \(f\left( x \right)\) at \({x_0}:\)

\[

{\lim\limits_{\Delta x \to 0} \Delta y = 0,\;\;}\Rightarrow

{\lim\limits_{\Delta x \to 0} \left| {M{M_1}} \right| }

= {\lim\limits_{\Delta x \to 0} \sqrt {{{\left( {\Delta x} \right)}^2} + {{\left( {\Delta y} \right)}^2}} = 0.}

\]

The limiting position of the secant \(M{M_1}\) is just the tangent line to the graph of the function \(y = f\left( x \right)\) at point \(M.\)

There are two kinds of tangent lines – oblique (slant) tangents and vertical tangents.

#### Definition \(1\).

If there is a finite limit \(\lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) = {k_0},\) then the straight line given by the equation

\[y – {y_0} = k\left( {x – {x_0}} \right),\]

is called the oblique (slant) tangent to the graph of the function \(y = f\left( x \right)\) at the point \(\left( {{x_0},{y_0}} \right).\)

#### Definition \(2\).

If the limit value of \(k\) as \(\Delta x \to 0\) is infinite: \(\lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) = \pm \infty,\) then the straight line given by the equation

\[x = {x_0},\]

is called the vertical tangent to the graph of the function \(y = f\left( x \right)\) at the point \(\left( {{x_0},{y_0}} \right).\)

It is important that

\[

{{k_0} = \lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) }

= {\lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} }

= {f’\left( {{x_0}} \right),}

\]

that is the slope of the tangent line is equal to the derivative of the function \(f\left( {{x_0}} \right)\) at the tangency point \({x_0}.\) Therefore, the equation of the oblique tangent can be written in the form

\[

{y – {y_0} = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right)\;\;\text{or}}\;\;\;\kern-0.3pt

{y = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right) + f\left( {{x_0}} \right).}

\]

Since the slope of a straight line is equal to the tangent of the slope angle \(\alpha,\) which the line forms with the positive direction of the \(x\)-axis, then the following triple identity is valid:

\[k = \tan \alpha = f’\left( {{x_0}} \right).\]

### Equation of a Normal Line in Cartesian Coordinates

A straight line perpendicular to the tangent and passing through the point of tangency \(\left( {{x_0},{y_0}} \right)\) is called the normal to the graph of the function \(y = f\left( x \right)\) at this point \(\left({\text{Figure }2}\right).\)

From geometry it is known that the product of the slopes of perpendicular lines is equal to \(-1.\) Therefore, knowing the equation of a tangent at the point \(\left( {{x_0},{y_0}} \right):\)

\[y – {y_0} = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right),\]

we can immediately write the equation of the normal in the form

\[y – {y_0} = – \frac{1}{{f’\left( {{x_0}} \right)}}\left( {x – {x_0}} \right).\]

#### Special Cases

- If the derivative \(f^\prime\left( {{x_0}} \right)\) is zero, then we have a horizontal tangent line. This means that the normal line at this point is a vertical line. It is defined by the equation \[{x = {x_0}.}\]
- If the derivative \(f^\prime\left( {{x_0}} \right)\) approaches (plus or minus) infinity, we have a vertical tangent. In this case, the normal line is a horizontal line defined by the equation \[{y = {y_0}.}\]

### Equations of Tangent and Normal Lines in Parametric Form

Let a plane curve be given parametrically:

\[{x = x\left( t \right),}\;\;\;\kern-0.3pt{y = y\left( t \right).}\]

Then the slope of the tangent drawn at the point \(\left( {{x_0},{y_0}} \right)\) can be found using the differentiation rule for parametric functions:

\[k = \tan \alpha = \frac{{{y’_t}}}{{{x’_t}}}.\]

The equation of the tangent is given by

\[

{y – {y_0} = \frac{{{y’_t}}}{{{x’_t}}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt

{\text{or}\;\;\;\kern-0.3pt\frac{{x – {x_0}}}{{{x’_t}}} = \frac{{y – {y_0}}}{{{y’_t}}}.}

\]

Accordingly, the equation of the normal is written as

\[

{y – {y_0} = – \frac{{{x’_t}}}{{{y’_t}}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt

{\text{or}\;\;\;\kern-0.3pt\frac{{x – {x_0}}}{{{y’_t}}} = – \frac{{y – {y_0}}}{{{x’_t}}}.}

\]

### Equations of Tangent and Normal Lines in Polar Coordinates

Suppose that a curve is defined by a polar equation \(r = f\left( \theta \right),\) which expresses the dependence of the length of the radius vector \(r\) on the polar angle \(\theta.\) In Cartesian coordinates, this curve will be described by the system of equations

\[\left\{ \begin{array}{l} x = r\cos \theta = f\left( \theta \right)\cos \theta \\ y = r\sin \theta = f\left( \theta \right)\sin\theta \end{array} \right..\]

Thus, we have written the parametric equation of the curve, where the angle \(\theta\) plays the role of a parameter.Next, it is easy to obtain an expression for the slope of the tangent to the curve at the point \(\left( {{x_0},{y_0}} \right):\)

\[ {k = \tan \theta = \frac{{{y’_\theta }}}{{{x’_\theta }}} } = {\frac{{{{\left( {r\sin \theta } \right)}^\prime }}}{{{{\left( {r\cos \theta } \right)}^\prime }}} } = {\frac{{{r’_\theta }\sin \theta + r\cos \theta }}{{{r’_\theta }\cos\theta – r\sin \theta }}.} \]

As a result, the equations of the tangent and normal lines are written as follows:

\[ {y – {y_0} = \frac{{{y’_\theta }}}{{{x’_\theta }}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {(\text{tangent}),} \]

\[ {y – {y_0} = -\frac{{{x’_\theta }}}{{{y’_\theta }}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {(\text{normal}).} \]

The study of curves can be performed directly in polar coordinates without transition to the Cartesian system. In this case, instead of the angle \(\theta\) with the polar axis (i.e. with the positive direction of the \(x\)-axis), it is more convenient to use the angle \(\beta\) with the line containing the radius vector \(r\) (Figure \(3\)).

The tangent of the angle \(\beta\) is calculated by the formula

\[\tan \beta = \frac{r}{{{r’_\theta }}}.\]

The angle formed by the normal and the extended radius vector is \(\beta + \large\frac{\pi }{2}\normalsize.\) Using the reduction identity, we get:

\[

{\tan \left( {\beta + \frac{\pi }{2}} \right) }

= { – \cot \beta = – \frac{1}{{\tan \beta }} }

= { – \frac{{{r’_\theta }}}{r}.}

\]

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the equation of the tangent to the curve \({y = \sqrt x}\) at the point \({\left({1,1}\right)}\) (Figure \(4\)).### Example 2

Find a point on the curve \(y = {x^2} – 2x – 3\) at which the tangent is parallel to the \(x-\)axis.### Example 3

Find the equation of the tangent line to the curve \(y = {x^4}\) at the point \(\left( { – 1,1} \right).\)### Example 4

Find the equation of the tangent line to the curve \(y = {x^3}\) at \({x_0} = 1.\)### Example 5

Find the equation of the tangent line to the curve \(y = \ln {x^2}\) that is parallel to the straight line \(y = x.\)### Example 6

Find a point on the curve \(y = \sqrt x, \) where the tangent makes an angle of 45 degrees with the positive \(x-\)axis.### Example 7

Find the equation of the normal line to the curve \(y = {x^3} + {e^x}\) at \({x_0} = 0.\)### Example 8

The equation of the tangent line to the graph of a function at \({x_0} = 1\) is defined by the equation \(2x + y – 4 = 0.\) Find the equation of the normal line passing through this point.### Example 9

Find the equation of the normal to the graph of the function \(y = \large{\frac{{x + 1}}{{x – 1}}}\normalsize\) at the point where \(x = 2.\)### Example 10

Find the equations of the tangent line and normal line to the parabola \(y = 2{x^2}\) at the point \(\left( {2,8} \right).\)### Example 11

Write an equation of the normal to the ellipse \[\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1\] at the point \(\left( {1,\large\frac{{\sqrt 3 }}{2}\normalsize} \right)\) (Figure \(5\)).### Example 12

Find the angle at which the parabola \(y = {x^2} – \frac{1}{4}\) intersects the positive \(x-\)axis.### Example 13

Find the angles at which the curve \(y = {x^3} – x\) intersects the \(x\)-axis.### Example 14

The normal drawn to the hyperbola \(y = \large{\frac{1}{x}}\normalsize\) at a point \({x_0}\) makes an angle of \(45^\circ\) with the positive \(x-\)axis. Determine the coordinate \({x_0}\) of the point of tangency (Figure \(6\)).### Example 15

Write equations of the tangent and normal to the graph of the function \(y = x\sqrt {x – 1} \) at \(x = 2.\)### Example 16

Find the equations of the tangent and normal lines to the graph of the natural logarithm function at \({x_0} = 1.\)### Example 17

Find the equation of the normal line to the curve \(y = {\mathop{\rm arccot}\nolimits} \large{\frac{1}{x}}\normalsize\) at \(x = 1.\)### Example 18

Given the parabola \(y = 2{x^2}.\) A secant is drawn through the \(2\) points of the parabola, which have the coordinates \(x = -1\) and \(x = 2\) (Figure \(8\)). Find the tangent to the parabola parallel to the secant.### Example 19

A tangent line is drawn to the graph of the function \(y = \large{\frac{1}{x}}\normalsize\) at the point \(\left( {1,1} \right)\) (Figure \(9\)). Find the length of the tangent segment \(AB\) in the first quadrant.### Example 20

The tangent and normal lines are drawn to the parabola \(y = {x^2}\) at the point \({x_0} = 2\) (Figure \(10\)). Find the length of the line segment \(AB\) between the points of intersection of the lines with the \(x-\)axis.### Example 21

Determine the area of the triangle formed by the tangent to the graph of the function \(y = 3 – {x^2}\) drawn at the point \(\left( {1,2} \right)\) and the coordinate axes (Figure \(11\)).### Example 22

The normal drawn to the curve \(y = \ln x\) at the point \(M\left( {1,0} \right)\) intersects the \(y-\)axis at the point \(A\) (Figure \(12\)). Find the area \(S\) of the triangle \(AOM.\)### Example 23

A parabola is defined by the equation \(y = {x^2} + 2x + 3.\) Write equations of the tangents to the parabola passing through the point \(A\left( { – 1,1} \right).\)### Example 24

Find the equation of the tangent line drawn to the curve \({y^4} – 4{x^4} – 6xy = 0\) at the point \(M\left( {1,2} \right).\)### Example 25

Find the equation of the normal line to the curve \({x^3} + {y^2} + 2x – 6 = 0\) at the point \(\left( { – 1,3} \right).\)### Example 26

Prove that the curves \({x^2} – {y^2} = 3\) and \(xy = 2\) intersect at the right angle.### Example 27

At what points do the curves \(y = {x^2}\) and \(y = \sqrt x \) intersect? Find the angles of intersection between the curves at these points.### Example 28

Find the distance between the origin and the normal to the curve \(y = {e^{2x}} + {x^2}\) at the point where \(x = 0\) (Figure \(16\)).### Example 29

Find the angle between the tangent to the cardioid \(r = a\left( {1 + \cos \theta } \right)\) and the radius vector of the point of tangency (Figure \(17\)).### Example 30

Find equation of the tangent and normal to the astroid \(x = a\,{\cos ^3}t,\) \(y = a\,{\sin ^3}t\) at the point \(t = \large\frac{\pi }{4}\normalsize\) (Figure \(18\)).### Example 31

A tangent is drawn to the graph of the function \(y =\cos x\) at the point \(M\left( {{x_0},{y_0}} \right),\) where \(0 \lt {x_0} \lt \large\frac{\pi }{2}\normalsize\) (Figure \(19\)). Find the value of \({x_0},\) at which the area of the triangle formed by the tangent and the coordinate axes will be the least.### Example 1.

Find the equation of the tangent to the curve \({y = \sqrt x}\) at the point \({\left({1,1}\right)}\) (Figure \(4\)).Solution.

\[ {y’ = f’\left( x \right) } = {{{\left( {\sqrt x } \right)’ } = {\frac{1}{{2\sqrt x }}}, }}\\ {{f’\left( {{x_0}} \right) = f’\left( 1 \right) } = {\frac{1}{{2\sqrt 1 }} = \frac{1}{2}},}\\ {{{x_0} = 1,\;{y_0} = 1,\;\;}\kern0pt {f’\left( {{x_0}} \right) = \frac{1}{2}}} \]

Substitute the \(3\) values into the equation of the tangent line:

\[ {{{y – {y_0} } = {f’\left( {{x_0}} \right)\left( {x – {x_0}} \right)} }}. \]

This yields:

\[{y – 1 = \frac{1}{2}\left( {x – 1} \right)\;\;}\Rightarrow {y – 1 = \frac{x}{2} – \frac{1}{2}\;\;}\Rightarrow {y = \frac{x}{2} – \frac{1}{2} + 1\;\;}\Rightarrow {{y = \frac{x}{2} + \frac{1}{2}} } .\]

Answer:

\[{y = \frac{x}{2} + \frac{1}{2}}.\]

### Example 2.

Find a point on the curve \(y = {x^2} – 2x – 3\) at which the tangent is parallel to the \(x-\)axis.Solution.

Since the tangent is parallel to the \(x-\)axis, the derivative is equal to zero at this point. Hence,

\[y^\prime = \left( {{x^2} – 2x – 3} \right)^\prime = 2x – 2 = 0.\]

Then we find that

\[{x_0} = 1.\]

### Example 3.

Find the equation of the tangent line to the curve \(y = {x^4}\) at the point \(\left( { – 1,1} \right).\)Solution.

First we find the derivative of the function:

\[f’\left( x \right) = \left( {{x^4}} \right)’ = 4{x^3}.\]

Calculate the value of the derivative at \({x_0} = – 1:\)

\[{f’\left( {{x_0}} \right) = f’\left( { – 1} \right) }={ 4 \cdot {\left( { – 1} \right)^3} }={ – 4.}\]

Substitute the \(3\) known numbers and find the equation of the tangent line:

\[y – {y_0} = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right),\]

\[y – 1 = – 4\left( {x – \left( { – 1} \right)} \right),\]

\[y – 1 = – 4\left( {x + 1} \right),\]

\[y – 1 = – 4x – 4,\]

\[y = – 4x – 3.\]

### Example 4.

Find the equation of the tangent line to the curve \(y = {x^3}\) at \({x_0} = 1.\)Solution.

First we find the derivative:

\[{y^\prime = f^\prime\left( x \right) }={ \left( {{x^3}} \right)^\prime }={ 3{x^2}.}\]

The value of the derivative at the point of tangency is

\[f^\prime\left( {{x_0}} \right) = 3 \cdot {1^2} = 3.\]

Calculate \({y_0}:\)

\[{y_0} = {\left( {{x_0}} \right)^3} = {1^3} = 1.\]

Substitute this in the equation of tangent:

\[y – 1 = 3\left( {x – 1} \right),\]

\[y – 1 = 3x – 3,\]

\[y = 3x – 2.\]

### Example 5.

Find the equation of the tangent line to the curve \(y = \ln {x^2}\) that is parallel to the straight line \(y = x.\)Solution.

The derivative of the function is given by

\[{y’ = \left( {\ln {x^2}} \right)’ }={ \frac{1}{{{x^2}}} \cdot 2x }={ \frac{2}{x}.}\]

The slope of the tangent line must be equal to \(1\) as it follows from the equation of the straight line. This allows to find the tangency point:

\[\frac{2}{x} = 1, \Rightarrow {x_0} = 2.\]

Calculate the value of the function at this point:

\[{y_0} = y\left( 2 \right) = \ln {2^2} = \ln 4.\]

Now we can write the equation of the tangent line:

\[y – {y_0} = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right),\]

\[y – \ln 4 = 1 \cdot \left( {x – 2} \right),\]

\[y – \ln 4 = x – 2,\]

\[y = x + \ln 4 – 2.\]

### Example 6.

Find a point on the curve \(y = \sqrt x, \) where the tangent makes an angle of 45 degrees with the positive \(x-\)axis.Solution.

We use the triple identity

\[k = \tan \alpha = f’\left( {{x_0}} \right).\]

This yields

\[k = \tan 45^\circ = 1,\]

so the derivative is equal to

\[f’\left( {{x_0}} \right) = 1.\]

From the other side,

\[{\left( {\sqrt x } \right)’ = \frac{1}{{2\sqrt x }},}\;\; \Rightarrow {\frac{1}{{2\sqrt {{x_0}} }} = 1.}\]

Therefore

\[{2\sqrt {{x_0}} = 1,}\;\;\Rightarrow{\sqrt {{x_0}} = \frac{1}{2},}\;\;\Rightarrow{{x_0} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}.}\]

### Example 7.

Find the equation of the normal line to the curve \(y = {x^3} + {e^x}\) at \({x_0} = 0.\)Solution.

Determine the value of the function at \({x_0} = 0.\)

\[{y_0} = y\left( 0 \right) = {0^3} + {e^0} = 1.\]

The derivative is given by

\[{y^\prime\left( x \right) = \left( {{x^3} + {e^x}} \right)^\prime }={ 3{x^2} + {e^x}.}\]

At the point \({x_0} = 0,\) it equals

\[y^\prime\left( 0 \right) = 3 \cdot {0^2} + {e^0} = 1.\]

Thus, the equation of the normal is written as follows:

\[y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\]

\[y – 1 = – \frac{1}{1}\left( {x – 0} \right),\]

\[y = – x + 1.\]

### Example 8.

The equation of the tangent line to the graph of a function at \({x_0} = 1\) is defined by the equation \(2x + y – 4 = 0.\) Find the equation of the normal line passing through this point.Solution.

We rewrite the equation of the tangent as

\[y = – 2x + 4\]

and find the \(y-\)coordinate of the tangency point:

\[{y_0} = – 2 \cdot 1 + 4 = 2.\]

The slope of the tangent line is \(-2.\) Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, we get that the slope of the normal is equal to \(\large{\frac{1}{2}}\normalsize .\) So the equation of the normal can be written as

\[y – {y_0} = k\left( {x – {x_0}} \right),\]

\[y – 2 = \frac{1}{2}\left( {x – 1} \right),\]

\[y – 2 = \frac{x}{2} – \frac{1}{2},\]

\[2y – 4 = x – 1,\]

or

\[x – 2y + 3 = 0.\]

### Example 9.

Find the equation of the normal to the graph of the function \(y = \large{\frac{{x + 1}}{{x – 1}}}\normalsize\) at the point where \(x = 2.\)Solution.

Differentiate the given function using the quotient rule:

\[{\require{cancel}y^\prime = \left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime }={ \frac{{\left( {x + 1} \right)^\prime\left( {x – 1} \right) – \left( {x + 1} \right)\left( {x – 1} \right)^\prime}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{x – 1 – \left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{\cancel{x} – 1 – \cancel{x} + 1}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{ – 2}}{{{{\left( {x – 1} \right)}^2}}}.}\]

At the point \({x_0} = 2,\) the function and the derivative have the following values:

\[y\left( 2 \right) = \frac{{2 + 1}}{{2 – 1}} = 3,\]

\[{y^\prime\left( 2 \right) = f^\prime\left( 2 \right) }={ – \frac{2}{{{{\left( {2 – 1} \right)}^2}}} = – 2.}\]

So the equation of the normal is given by

\[y – {y_0} = – \frac{1}{{f^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\]

\[y – 3 = – \frac{1}{{\left( { – 2} \right)}}\left( {x – 2} \right),\]

\[y – 3 = \frac{x}{2} – 1,\]

\[y = \frac{x}{2} + 2.\]

### Example 10.

Find the equations of the tangent line and normal line to the parabola \(y = 2{x^2}\) at the point \(\left( {2,8} \right).\)Solution.

Calculate the derivative of the function:

\[{y^\prime = \left( {2{x^2}} \right)^\prime = 4x,}\;\; \Rightarrow {y^\prime\left( 2 \right) = 8.}\]

Write the equation of the tangent line:

\[y – {y_0} = y^\prime\left( {{x_0}} \right)\left( {x – {x_0}} \right),\]

\[y – 8 = 8\left( {x – 2} \right),\]

\[y – 8 = 8x – 16,\]

\[8x – y – 8 = 0.\]

Similarly, we get the equation of the normal line:

\[y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\]

\[y – 8 = – \frac{1}{8}\left( {x – 2} \right),\]

\[y – 8 = – \frac{x}{8} + \frac{1}{4},\]

\[8y – 64 = – x + 2,\]

\[x + 8y – 66 = 0.\]

So the answer is given by

\[{8x – y – 8 = 0,\;\;}\kern0pt{x + 8y – 66 = 0.}\]