## Equation of a Tangent Line: Problems and Solutions

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Anonymous
1 month ago

Find the equation(s) of any normals to the curve y = x + 1/x at points where the tangent
is parallel to the line y = -3x.

Matheno
Editor
1 month ago

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! : )

Ferro Baltazar
6 months ago

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

Matheno
Editor
6 months ago

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! : )

Anonymous
1 year ago

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

Matheno
Editor
1 year ago

(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.

Anonymous
2 years ago

Y=x³at the point 1,1 give the solution

Matheno
Editor
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):
$$y – 1 = 3(x-1)$$