Equation of a Tangent Line: Problems and Solutions

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Anonymous
3 days ago

for the curve y=x^2 + x at what point does the normal line at (0,0) intersect the tangent line at (1,2)?

Matheno
Editor
3 days ago

Our solution will have three big steps: (1) Find the equation of the normal line at (0, 0). (2) Find the equation of the tangent line at (1, 2). (3) Find where those two lines intersect.

(1) As explained above, since we have f'(0) = 1, the normal line has slope m_normal = -1. Hence the equation of this normal line is [y – 0 = -(x -0)], or just y = -x.

(2) We have f'(1) = 3, and so the slope of the tangent line is m_tangent = 3. The equation of this tangent line is thus y -2 = 3(x-1), or y = 3x + 1.

(3) Then setting the equations for those two lines equal to each other, we find x = 1/4. (We’ll leave that algebra to you.)

We hope that all makes sense and is helpful!

Anonymous
1 month ago

F(x)=4x^2-4x+1. find the points of tangent where it is perpendicular to the line 2y+x-4=0

Last edited 1 month ago by Anonymous
Matheno
Editor
1 month ago

Great question! We’re going to assume that there’s an equals-sign at the end of the line’s equation, so “…perpendicular to the line $2y+x-4=0$.” If that’s not right, we trust you’ll be able to follow the solution steps and adjust accordingly.

Our solution has three big steps: (1) First, we need the slope of that line. (2) Determine the slope of lines perpendicular to that one. (3) Find points on the function’s curve that have tangent lines with that same slope.

Step 1: The line’s slope is easy to see if we rewrite the equation in Point-Slope form:
\begin{align*}
2y+x-4 &= 0 \8px] 2y &= -x +4 \\[8px] y &= -\dfrac{1}{2} x + 2 \quad \blacktriangleleft \\[8px] \end{align*} Now that the equation for the line is written in Point-Slope form, y = mx + b, we can readily see the slope is m_{\text{line}}=-\dfrac{1}{2}. Step 2: We want our tangent lines to be perpendicular to this line, so we want them to have slope m_\text{tangent} = -\dfrac{1}{m_\text{line}}: \[ m_{\text{tangent}}= – \dfrac{1}{- \frac{1}{2}}
$m_{\text{tangent}} = 2 \quad \blacktriangleleft$
Step 3: Now we need to find where $F(x)$ has slope $m_{\text{tangent}}=2:$ We first need the function’s derivative:
\begin{align*}
F(x) &= 4x^2 – 4x + 1 \8px] F'(x) &= 8x – 4 \quad \blacktriangleleft \\[8px] \end{align*} Since F'(x) is the slope of F(x), we set F'(x)=2 and solve for x. \begin{align*} F'(x) &= 8x – 4 \\[8px] 8x-4 &= 2 \\[8px] 8x &= 6 \\[8px] x &= \dfrac{3}{4} \quad \blacktriangleleft \\[8px] \end{align*} That’s the x-value of the one point where the tangent has the slope we’re after. Now let’s find its y-value: \begin{align*} F \left( \dfrac{3}{4} \right) &= 4 \left( \dfrac{3}{4} \right)^2 – 4 \left( \dfrac{3}{4} \right) + 1 \\[8px] &=4 \left( \dfrac{9}{16} \right) – 4 \left( \dfrac{3}{4} \right)+1 \\[8px] &= \dfrac{9}{4} – 3 + 1 \\[8px] &= \dfrac{9}{4} – 2 \\[8px] &= \dfrac{1}{4} \quad \blacktriangleleft \\[8px] \end{align*} Finally, the point where F(x) is tangent to the line perpendicular to 2y + x -4 = 0 is: \[ ( 3/4, 1/4) \cmark
We hope that helps!

Last edited 1 month ago by Matheno
Anonymous
6 months 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
6 months 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
10 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
10 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