## Equation of a Tangent Line: Problems and Solutions

### Equation of a Tangent Line or Normal Line: Problems and Solutions

Are you working to find the equation of a tangent line (or normal line) in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

To answer these questions, you will almost always use the Point-Slope form of a line. Recall that if a line has slope *m *and contains the point $(x_0, y_0)$, then you can write its equation as:

**Point-Slope form of a line:**

$$\bbox[yellow,5px]{y – y_0 = m(x – x_0)}$$

The problem statement typically specifies the point $(x_0, y_0)$, and so really these problems come down to determining the slope *m *of the line — which we’ll address below.

You will use that equation again and again; memorize it if you don’t know it already.

(It’s just a variant on the definition of slope: $m = \dfrac{y – y_0}{x – x_0}.)$

**I. Tangent Line to a Curve**

*Very* frequently in beginning Calculus you will be asked to find an equation for the line **tangent** to a curve at a particular point. We’re calling that point $(x_0, y_0)$.

To find the line’s equation, you just need to remember that the tangent line to the curve has slope equal to the derivative of the function evaluated at the point of interest:

$$\bbox[yellow,5px]{m_\text{tangent line} = f'(x_0)}$$

That is, find the derivative of the function $f'(x)$, and then evaluate it at $x = x_0$. That value, $f'(x_0),$ *is *the slope of the tangent line.

Hence we can write the equation for the tangent line at $(x_0, y_0)$ as

\[\bbox[10px,border:2px solid blue]{

\begin{align*}

y – y_0 &= m_\text{tangent line}(x – x_0) \\[8px]
y – y_0 &= f'(x_0)(x – x_0)

\end{align*} } \]

If those equations look abstract to you, don’t worry. As soon as you work a few problems, the process will make sense — we promise.

**II. Normal Line to a Curve**

Sometimes instead a question will ask you instead to find the line **normal** to a curve. That’s the same thing as asking for the line that is **perpendicular** to the curve.

You will again use the Point-Slope form of a line. But now to compute the slope of the line, recall that the slopes of perpendicular lines are the negative reciprocals of each other ($m_2 = -\dfrac{1}{m_1}$). We want the slope of the line that is perpendicular to the curve at a point, and hence that is perpendicular to the tangent line to the curve at that point:

\[\bbox[yellow,5px]{

\begin{align*}

m_\text{normal line} &= \frac{-1}{m_\text{tangent line}}\\[12px]
&= \frac{-1}{f'(x_0)}

\end{align*}}\]

Hence we can write the equation for the normal line at $(x_0, y_0)$ as

\[\bbox[10px,border:2px solid blue]{

\begin{align*}

y – y_0 &= m_\text{normal line}(x – x_0) \\[8px]
y – y_0 &= \frac{-1}{f'(x_0)}(x – x_0)

\end{align*} } \]

We recommend *not* trying to memorize all of the formulas above. Instead, remember the Point-Slope form of a line, and then use what you know about the derivative telling you the slope of the tangent line at a given point. The problems below illustrate.

Problem 1 illustrates the process of putting together different pieces of information to find the equation of a tangent line.

Problem 2 requires that you find the pieces of information before you can put them together.

Since $g'(5) = 2$, we know that the slope of the tangent line at $x = 5$ is $m=2$. Furthermore, the tangent line contains the point (5, -3), since it passes through (grazes) that point on the curve.

Then using the point-slope form of a line that contains the point $(x_o, y_o)$ we have

\begin{align*}

y – y_o &= m(x – x_o) \\

y – (-3) &= 2(x-5) \\

y + 3 &= 2(x-5) \quad \cmark

\end{align*}

We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form:

\begin{align*}

y+3 &= 2x – 10 \\

y &= 2x – 10 -3 \\

y &= 2x – 13 \quad \cmark

\end{align*}

Either answer is valid and correct.

**(a)**Find the equation to the line tangent to the curve at the point (1, 5).

**(b)**Find the equation of the line normal (perpendicular) to the curve at the point (1, 5).

Hence we just need its slope *m*, which is is the same as the slope of the curve at that point $(x=1).$ And that slope equals the function’s derivative at that point:

\begin{align*}

f(x) &= x^3 – x + 5\\[8px]
f'(x) &= 3x^2 -1 \\[8px]
f'(1) &= 3(1) – 1 = 2

\end{align*}

Hence the tangent line has slope $m_\text{tangent} = f'(1) = 2.$

Then we can write the equation of this tangent line, using the point-slope form of a line that has slope $m_\text{tangent} = 2$ and contains the point $(1, 5)$:

\begin{align*}

y – y_o &= m_\text{tangent}(x – x_o) \\[8px]
y – 5 &= (2)(x – 1) \quad \cmark

\end{align*}

We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form:

\begin{align*}

y – 5 &= 2x -2 \\[8px]
y &= 2x – 2 + 5 \\[8px]
y&= 2x +3 \quad \cmark

\end{align*}

Either answer is valid and correct.

$$m_2 = -\frac{1}{m_1}$$

From part (a) we know that the line tangent to the curve at the point (1,5) has slope $m_\text{tangent} = 2$, and so our perpendicular line at that point has slope

\begin{align*}

m_\text{normal} &= -\frac{1}{m_\text{tangent}} \\[8px] &= -\frac{1}{2}

\end{align*}

Hence the normal line has slope $m_\text{normal} = -\dfrac{1}{2}.$

This line passes through the curve at point (1, 5) and has slope $m_\text{normal} = -\dfrac{1}{2}$, and so its equation is given by

\begin{align*}

y – y_o &= m_\text{normal}(x – x_o) \\[8px]
y-5 &= -\dfrac{1}{2} (x -1) \quad \cmark

\end{align*}

We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form:

\begin{align*}

y – 5 &= -\dfrac{1}{2}x + \dfrac{1}{2} \\[8px]
y &= -\dfrac{1}{2}x + \dfrac{1}{2} + 5 \\[8px]
y &= -\dfrac{1}{2} x + \frac{11}{2} \quad \cmark

\end{align*}

Either answer is valid and correct.

*[This is a more challenging problem, submitted by a student in the comments below.]*

A line normal (perpendicular) to the curve $y = 2x^2$ at a point in the first quadrant also passes through the point $\left(0, \dfrac{3}{4}\right)$. Find an equation for this line.

As usual, a quick figure helps a lot. Here we’ve shown the curve $y = x^2,$ and the line that passes through the point $(0, \frac{3}{4})$. This line intersects the curve at the point we’re calling $(x_1, y_1).$

**Note that naming this point on the curve with coordinates like this is crucial to our solution.**

There are several pieces of information we have to put together to solve this problem.

(1) The first is that the slope of a line that is normal (perpendicular) to this curve at the point $(x_1, y_1)$ is given by

\[m_\text{normal, at $x=x_1$} = -\dfrac{1}{f'(x_1)} \]
Since $f(x) = 2x^2,$ we have $f'(x) = 4x.$

For the particular point of interest, $(x_1, y_1),$ where the line intersects the curve, we can write the slope of the normal line as

\[m_\text{normal, at $x=x_1$} = -\dfrac{1}{4x_1} \quad \triangleleft \quad (1) \]
Keep that in mind for a moment.

(2) We *also* know that the line contains the points $(0, \frac{3}{4})$ and $(x_1, y_1).$ Hence we can write its slope as

\[m_\text{line} = \frac{y_1-\frac{3}{4}}{x_1-0} \quad \triangleleft \quad (2) \]
Now, the magic: the line must actually meet both conditions (1) and (2), and so we must have

\[m_\text{normal, at $x=x_1$} = m_\text{line}\]
This requirement will let us solve for $y_1,$ as you’ll see:

\begin{align*}

m_\text{normal, at $x=x_1$} &= m_\text{line} \\[8px]
-\dfrac{1}{4x_1} &= \frac{y_1-\frac{3}{4}}{x_1} \\[8px]
-\dfrac{1}{4} &= y_1-\frac{3}{4} \\[8px]
– y_1 &= -\frac{3}{4} + \dfrac{1}{4} \\[8px]
-y_1 &= -\frac{1}{2} \\[8px]
y_1 &= \frac{1}{2} \quad \triangleleft

\end{align*}

We now have the *y*-value where the line intersects the curve. That leaves us with a few steps to go, since the question asked for the equation of the line. So let’s next determine $x_1$:

We know that the point $(x_1, y_1)$ lies on the curve $y = 2x^2,$ and so since $y_1 = \frac{1}{2}$ we must have

\begin{align*}

\frac{1}{2} &= 2x_1^2 \\[8px]
x_1^2 &= \frac{1}{4} \\[8px]
x_1 &= \pm \sqrt{\frac{1}{4}} \\[8px]
&= \pm \frac{1}{2}

\end{align*}

Ah, but the problem specifies “at a point in the first quadrant,” and so we choose the *positive* solution:

\[x_1 = \frac{1}{2} \quad \triangleleft \]
And then we immediately know the slope of the normal line, since we decided (1) above

\[m_\text{normal, at $x=x_1$} = -\dfrac{1}{4x_1} \]
we must have

\begin{align*}

m_\text{normal, at $x=1/2$} &= -\dfrac{1}{4\left(\frac{1}{2} \right)} \\[8px]
&= – \frac{1}{2} \quad \triangleleft

\end{align*}

Now we’re actually almost done: we know that the slope of the normal line is $m_\text{normal} = -\dfrac{1}{2},$ and since we were told the line passes through the point $(0, \frac{3}{4})$ we know it has *y*-intercept $b = \frac{3}{4}.$ Hence using the slope-intercept form of a line:

\begin{align*}

y &= m_\text{normal}x + b \\[8px]
y &= -\frac{1}{2}x + \frac{3}{4} \quad \cmark

\end{align*}

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Find the equation(s) of any normals to the curveat points where the tangent

y= x + 1/xis parallel to the line

y = -3x.Let’s break this down:

Step 1: We need ” points where the tangent is parallel to the line y = -3x.” Since that line has slope equal to -3, we’re looking for points on the curve where the tangent line has that slope.And since the slope of the tangent line to the curve at the point $(a, f(a))$ has slope $m_\text{tangent} = f'(a),$ we need to identify the points where $f'(a) = -3.$ So let’s find those:

\begin{align*}

f(x) &= x + \frac{1}{x} \\[8px]

f'(x) &= 1 – \frac{1}{x^2} \\[8px]

\end{align*}

That’s true for any point . We want the particular points where $f'(a) = -3$:

\begin{align*}

f'(a) = 1 – \frac{1}{a^2} &= -3 \\[8px]

– \frac{1}{a^2} &= -4 \\[8px]

a^2 &= \frac{1}{4} \\[8px]

a &= \pm \frac{1}{2} \quad \blacktriangleleft

\end{align*}

Hence we have two points where the tangent line has slope equal to -3: $x_1 = -\dfrac{1}{2}$ and $x_2 = \dfrac{1}{2}$. See the top graph below.

Next, let’s find the equations of the lines that are normal (perpendicular) to the curve at these two points. To write the equation of a line, we need to know its slope, and one point on the line.

Now we already know the slope of these lines:

\[m_\text{normal} = -\frac{1}{m_\text{tangent}} = -\frac{1}{(-3)} = \frac{1}{3} \quad \blacktriangleleft\]

So, we just need both the x-value and y-value of one point on the line. We know from above the x-value of one point on each line, so we just need to find the corresponding y-value of those two points:

We know from above the x-value of one point on each line, so we just need to find the corresponding y-value of those two points:

Recall the original function

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

Hence for $x_1 = -\dfrac{1}{2}$, we have

\[ y_1 = -\frac{1}{2} + (-2) = -\frac{5}{2} \quad \blacktriangleleft\]

and so the point $(-\dfrac{1}{2}, -\dfrac{5}{2})$ is on the line and its equation, in point-slope form, is

\begin{align*}

y – \left(- \frac{5}{2}\right) &= \frac{1}{3}\left[ x – \left(- \frac{1}{2}\right)\right] \\[8px]

y + \frac{5}{2} &= \frac{1}{3}\left( x + \frac{1}{2}\right) \quad \cmark

\end{align*}

And for $x_2 = \dfrac{1}{2}$ we have

\[y_2 = \frac{1}{2} + 2 = \frac{5}{2} \quad \blacktriangleleft\]

and so the point $(\dfrac{1}{2}, \dfrac{5}{2})$ is on the line and its equation, in point-slope form, is

\[ y – \frac{5}{2} = \frac{1}{3}\left( x – \frac{1}{2}\right) \quad \cmark\]

Our final answer is the two equations with the green checkmarks next to them, and we’ve plotted those two lines, and the orignal function’s curve, on the lower graph below. BUT, much more important is the process we used to find those equations, and we hope that makes sense.

And we hope to have helped! : )

Normal line to the curve y=2x² at a point in the first quadrant passed through the point (0,¾) ,find an equation of this normal line please show solution

Hi, Ferro,

Thanks for asking! This is a good problem. We like it so much we added it as a sample problem #3 above.

We hope that helps! : )

Y=x5 at the point 1,5 give the solution

Thanks for asking! Would you please clarify? Either

(1) the given equation is y = 5x, in which case the function is a line and so is tangent to itself,

or

(2) the given equation is y = x^5, in which case the point (1,5) isn’t on the curve, but the point (1,1) is.

If you clarify which choice, if either, is the correct question, we’ll try to provide more detail about the solution.

Y=x³at the point 1,1 give the solution

Following the approach outlined in the SUMMARY box above:

1. The slope of the tangent line at (1,1) equals the derivative at that point. Since $y=f(x)= x^3,$ the derivative is $y’=f'(x) =3x^2.$ Hence the slope of the tangent line at $x_0 = 1$ is $m_{tangent} = f'(1) = 3.$

2. Then we can write the equation of this tangent line, using the point-slope form of the line that has slope $m_{tangent} = 3$ and contains the point (1,1):

$$ y – 1 = 3(x-1)$$

Does that provide the solution you were after? We hope so!