On this page we look at how to find the equation of a tangent and also the normal to a curve. A tangent is a straight line that touches a curve at one point and has the same gradient as the curve at that point. A normal is straight line that is perpendicular to the tangent to the curve. We use the gradient function and need to know a point on the line.
On this page, you should learn about
- equations of a tangent to a curve at a given point
- equations of a normal to a curve at a given point
The following videos will help you understand all the concepts from this page
A tangent is a straight line that touches a curve at one and only one point.
In the video below we are going to look at tangents to curves and how we can use differentiation to find the gradient to a tangent and therefore the equation of a tangent. We will look at the example
Find the equation of the tangent to the curve y = x² at the point P(3, 9)
Notes from the video
A normal line to a curve is the line that is perpendicular to the tangent to a curve at a particular point.
In the following video we are going to look at normals to curves, how we can use differentiation to find the gradient of a normal and therefore how to find the equation of a normal. In the example, we will also look at how to find the intersection of a normal with the curve
- Find the equation of the normal to the curve \(f(x)= \frac{1}{x}\)at the point P\((-2,-\frac{1}{2})\)
- Find the co-ordinates of the point where this normal meets the curve again.
Notes from the video
Print from here
Before you attempt to answer questions about equations of tangents and normals, you should be confident with finding the equation of a straight line and perpendicular lines. Here is a quiz that practices that knowledge
START QUIZ!
What is the gradient and y intercept of the following line?
gradient =
y intercept =
gradient = \(\frac{rise}{run}\)
The equation of the following line can be written in the form \(y=ax+b\)
Find a and b
a =
b =
a = gradient
b = y intercept
Which of the following is the correct equation of the following straight line
gradient = 3
y intercept = -1
Which of the following lines passes through the point (2 , 1)
y = -x + 3
1 = -2 + 3
Which of the following lines is parallel to \(y = \frac{1}{2}x+2\)
x - 2y + 3 = 0 can be rearranged into the form y = mx + c
x + 3 = 2y
\(y = \frac{1}{2}x+\frac{3}{2}\)
Which of he following gradients could represent the gradients of 2 perpendicular lines
A) 2
B) \(\frac{2}{5}\)
C) 5
D) \(-\frac{5}{2}\)
If a line has gradient m then a perpendicular line has gradient \(\frac{-1}{m}\)
What is the gradient of the line that is perpendicular to the line below
gradient =
The gradient a line perpendicular to \(y = -\frac{2}{3}x-2\) can be written in the form \(\frac{a}{b}\)
Find a and b
a =
b =
If a line has gradient m then a perpendicular line has gradient \(\frac{-1}{m}\)
The equation line below can be written in the form \(ax+by+c=0\)
Find a , b and c
a =
b =
c =
gradient = \(-\frac{2}{5}\)
y intercept = -2
\(y = -\frac{2}{5}x-2\\ 5y=-2x-10\\ 5y+2x+10=0\\ 2x+5y+10=0\)
The equation of the line that is perpendicular to the line below and that passes through the point (1,1) can be written in the form \(ax+2y+b=0\)
Find a and b
a =
b =
gradient of line in graph = \(-\frac{2}{5}\)
gradient of perpendicular line = \(\frac{5}{2}\)
\(y = \frac{5}{2}x+c \)
passes through (1,1)
\(1 = \frac{5}{2}\cdot1+c\\ -\frac{3}{2} = c \\ y = \frac{5}{2}x-\frac{3}{2} \\ 2y=5x-3\\ -5x+2y+3=0 \)
Here is a quiz about finding the equation of a tangent to a curve and the equation of a normal to a curve
START QUIZ!
f(x) = x3+2x
Find the gradient of the curve y = f(x) at the point (-1,-3)
gradient =
f'(x)=3x²+2
f '(-1) = 3(-1)²+2 = 5
The gradient of the curve \(y = \frac{1}{x^2}\) at point A is \(\frac{1}{4}\). Find the x coordinate at A
x =
y = x-2
\(\frac{dy}{dx}\)=-2x-3
\(\frac{-2}{x^3}=\frac{1}{4}\)
-8 = x3
x = -2
The gradient of the tangent below is \(\frac {a}{4}\) . Find a
a =
\(f(x)=x+\sqrt x\)
\(f(x)=x+{ x }^{ \frac { 1 }{ 2 } }\)
\(f'(x)=1+{ \frac { 1 }{ 2 } x }^{ -\frac { 1 }{ 2 } }\)
\(f'(x)=1+{ \frac { 1 }{ 2 \sqrt x }} \)
\(f'(4)=1+{ \frac { 1 }{ 2 \sqrt 4 }} \)
\(f'(4)=1+{ \frac { 1 }{ 4 }} \)
The equation of the tangent to the curve at a point P is given by \(y=-\frac { 1 }{ 2 } x+5\)
What is the gradient to the normal at the point P
Gradient =
The normal is perpendicular to the tangent.
\(gradient\ of \ normal\ =\frac { -1 }{ gradient\ of\ tangent } \)
The gradient of the normal to the curve y = 2x(x - 1) at the point (2,4) is
gradient =
y = 2x²-2x
\(\frac { dy }{ dx } =4{ x }-2\)
when x = 2 , \(\frac { dy }{ dx }= 4 \cdot2-2=6\)
The equation of the tangent to the curve at A is given by y = m x + c
Find m and c
m =
c =
The gradient of the tangent = 6
y = 6x + c
Tangent passes through (1,-2)
-2=6(1)+c
c=-8
The gradient of the tangent to the curve at (1,-3) is -2
The equation of the normal to the curve at A is given by x + ay + b = 0
Find a and b
a =
b =
gradient of tangent = -2
gradient of normal = \(\frac{1}{2}\)
y = \(\frac{1}{2}\)x + c
normal passes through the point (1,-3)
-3 = \(\frac{1}{2}\)+c
c = \(3\frac{1}{2}=\frac{7}{2}\)
y = \(\frac{1}{2}\)x + \(\frac{7}{2}\)
2y = x + 7
0 = x - 2y + 7
The equation of the tangent to the curve f(x)=x3 at the point (1,1) is given by y = mx -2
Find m
m =
f(x) = x3
f '(x) = 3x²
f '(1) = 3
Gradient of the tangent = 3
The equation of the normal to the curve f(x) = \(\sqrt x\) at the point (9,3) is given by y = mx + 57
Find m
m =
f(x) = \(\sqrt x\) =\(x^{\frac{1}{2}}\)
f '(x) = \(\frac{1}{2}x^{-\frac{1}{2}}\) = \(\frac{1}{2 \sqrt{x}}\)
f '(9) =\(\frac{1}{2 \sqrt{9}}=\frac{1}{6}\)
gradient of tangent = \(\frac{1}{6}\)
gradient of normal = -6
The equation of the normal to the curve y = \(\frac {2}{x}\) at (2,1) is y = 2x - 3
Find the coordinates (a,b) of the point where the normal crosses the curve again.
a =
b =
\(\frac {2}{x}=2x-3\)
2 = 2x² - 3x
2x² - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = \(-\frac {1}{2}\) , x = 2
when x = \(-\frac {1}{2}\), y = \(\frac {2}{-\frac{1}{2}}=-4\)
Let f(x) = (x - 1)(x - 4)(x + 2). The diagram below shows the graph of f and the point P where the graph crosses the x axis.
The line L is the tangent to the graph of f at the point P.
The line L intersects the graph of f at another point Q, as shown below
a) Find the coordinates of P
b) Show that \(f(x)=x^3-3x^2-6x+8\)
c) Find the equation of L in the form y = ax + b
d) Find the x coordinate of Q.
Hint
Full Solution
Let \(f(x)=\frac{x^4-4x^2}{4}\) .
C(2 , 0) lies on the graph of y = f(x)
a) The tangent to the graph of y = f(x) at C cuts the y axis at A. Find the coordinates of A.
b) The normal to the graph of y = f(x) at C cuts the y axis at B. Find the area of the triangle ABC.
Hint
Full Solution
The function \(f(x)=x^3-x^2-9x+9\) intersects the x axis at A, B and C.
The x coordinate of the point D is the mean of the x coordinates of B and C.
a) Find the coordinates of A, B and C.
b) Find the equation of the tangent to the curve at D.
c) Find the point of the intersection of the tangent with the curve. Interpret your result.
Hint
Full Solution
How much of Equation of Tangent and Normal have you understood?
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