4. TANGENTS AND NORMALS
§4.1 Equation of the Tangent at a Point
Recall that the slope of a curve at a point is the slope of the tangent at that point. The
slope of the tangent is the value of the derivative at that point. In computing the equation of
the tangent we have the slope plus one point, so we use the Slope Plus One Point Equation:
y  y0 = m(x x0) .
Example 1: Find the equation of the tangent to y = 3x2  4x + 7 at x = 1.
Solution: When x = 1, y = 6 so the tangent passes through (1, 6).
dy
dx = 6x  4. When x = 1 the slope is therefore 2.
The equation of the tangent is therefore y  6 = 2(x  1) = 2x  2, which simplifies to
y = 2x + 4.
One common error is to forget to substitute the particular value of x into the
derivative. If we’d made that mistake in the above example
we’d have concluded that the equation of the tangent is y  6
= (6x  4)(x  1) = 6x2  10x + 4, which simplifies to y = 6x2
 10x + 10. But this isn’t even the equation of a straight line.
Always make sure that the “m” in the equation is a constant –
you must substitute for x in the derivative.
TANGENT AT A POINT
Find the equation of the tangent to y = f(x) at P(x0, y0).
P(x0, y0)
(P lies on the curve.)
dy
(differentiate)
dx =
Slope of tangent is m =
(substitute x = x0).
It passes through (x0, y0) so equation of the tangent is:
y  y0 = m(x  x0) i.e. y 
=
(x 
)
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§4.2 Equation of the Normal at a Point
When a ray of light reflects off a curved surface, the
angle of reflection is the same as the angle of incidence.
A normal to a curve is a line that cuts the curve at right
angles. It’s therefore perpendicular to the tangent. So we use
the same y  y0 = m(x  x0) equation, but when we substitute
into the derivative to get the slope of the tangent we have to
invert and change the sign to get the perpendicular slope before
substituting for “m”.
Example 2: Find the equation of the normal to the curve y = x2 at x = 2.
Solution: When x = 2, y = 4 so the normal passes through
(2, 4).
dy
dx = 2x. So the slope of the tangent at x = 2 is 4. The slope
of the normal is therefore  ¼.
The equation of the tangent is therefore y  4 = ( ¼)(x  2)
=  ¼x + ½ .
This simplifies to x + 4y = 18.
NORMAL AT A POINT
Find the equation of the normal to y = f(x) at P(x0, y0).
P(x0, y0)
(P lies on the curve.)
dy
(differentiate)
dx =
Slope of tangent is
(substitute x = x0)
1
Therefore slope of the normal is: 
=
It passes through (x0, y0) so equation
y  y0 = m(x  x0) i.e.
y
=
of the normal is:
(x 
)
§4.3 Equation of Tangent(s) From a Point
Suppose we want the tangent or tangents that pass through some
point P that doesn’t lie on the curve. Because we don’t know the point of
contact we have to work indirectly.
Suppose the point of contact is Q, when x = t.
The variable “t” is called a “parameter”. It represents one or
Q
more specific values which we determine later. We now
work out the equation of the tangent in terms of “t”.
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Q
P
there may
several
possible Q’s
Now the fact that the tangent passes through the given point P means that we can
substitute the coordinates of P into this equation. This will give an equation in the parameter
“t” which we proceed to solve. Finally we feed each of these values of “t” into the equation of
the tangent and so get the final answer. (If this equation has no solutions this simply means
there are no tangents passing through the given point.)
Example 3: Find the equation of the tangents to y = x2 + x that pass through (1, 1).
Solution: Let the point of contact be when x = t. The corresponding y-value is y = t2 + t.
So the tangent passes through (t, t2 + t).
dy
dx = 2x + 1. The slope of he tangent at x = t is 2t + 1.
The equation of this tangent is therefore: y  (t2 + t) = (2t + 1)(x  t).
But since the tangent has to pass through (1, 1) we must have 1  (t2 + t) = (2t + 1)(1  t).
Simplifying this we get t2  2t = 0 and solving, we get t = 0 or t = 2.
When t = 0 the equation of the tangent is y  0 = 1(x  0), that is, y = x.
When t = 2 the equation of the tangent is y  (4 + 2) = 5(x  2), that is, y = 5x  4.
TANGENT FROM A POINT
Find the equation of the tangents to y = f(x) which pass through (x0, y0).
P(x0, y0)
(P does not lie on the curve.)
dy
(differentiate).
dx =
Slope of tangent is m =
(substitute x = t)
Passes through (x0, y0) so equation of tangent is:
y  f(t) = m(x  t) i.e.
y
=
(x  t)
substitute x = t in original equation
BUT it passes through (x0, y0) so
=
(put x = x0, y = y0 in equation of the tangent to get, after rearrangement
= 0 (an equation in t)
SOLVE this equation.
SUBSTITUTE each solution into
to get the required tangents.
EXERCISES FOR CHAPTER 4
Exercise 1: The slope of the tangent to a curve at P(2, 5) is 3. Write down the equation of the
tangent and the normal at P.
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Exercise 2: Find the equations of the tangents to the following curves at x = 4.
8
(i) y = x2  3x + 2;
(ii) y = x ;
1
(iii) y = x ;
(iv) y =
;
x3
(v) y = x x + 5 .
Exercise 3: Find the normals to the following curves at x = 2:
1
(i) y = 4 x4 + 1;
(ii) y = x + 2 ;
(iii) y = x x  1 .
Exercise 4: Find the tangents to y = x2  5x + 7 that pass through (2, 1).
1
Exercise 5: Find the tangents to y = x that pass through (2,  4).
Exercise 6: Find the tangents to y = x that pass through (1, 1).
[HINT: You will end up with an equation involving t . To solve this simply put T = t and
this will give you a quadratic in T. Once you have solved this for T, take square roots to get
the corresponding values for t. Only use positive square roots because t doesn’t exist for
negative t.]
Exercise 7: If y = 2x3  15x2 + 23x  10 find:
dy
(a) dx ;
(b) the points on this curve where the slope is 1;
(c) the tangents at these two points;
(d) (harder) the distance between these two parallel tangents.
Exercise 8: Find the equations of the three normals to y = x2 that pass through (0, 3/2).
[HINT: Proceed as for tangents passing through a point but use the slope of the normal instead
of the tangent.]
Exercise 9: A train moves along a track in the shape
of the parabola y = x2 (distances in kilometres). A
road runs along the x-axis and there’s a station at the
point where the track and the road come together.
You’re in a house 1 km from the station along this
road and, as the train approaches the station, the light
on the front of the train shines directly in the front
window. How many metres away is the train when
this occurs (to the nearest 100 metres)?
HINT: Let the train be at the point (t, t2).
Work out the equation of the tangent at this point.
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train
road
station house
Exercise 10: A very symmetrical man-made mound
rises from an otherwise flat plain. The vertical cross- light pole
sections of a hill through the highest point all have
the equation y = 9  x2 (with x giving the horizontal
distance in metres from the axis of symmetry and y,
also in metres, giving the height above the
surrounding plain). The horizontal cross-sections are
all circular. The mound casts a shadow around the
base of the plain. How far from the base of the
mound does this shadow reach?
mound
shadow
SOLUTIONS FOR CHAPTER 4
Exercise 1:
The equation of the tangent is y – y1 = m (x – x1), where m is the slope of the tangent and
(x0, y0) is the point on the curve.
So in this case we have: y – 5 = 3 (x – 2) which can be simplified to 3x – y – 5 = 0.
1
1
The slope of the normal is – 3 so the equation of the normal is y – 5 =  3 (x 2) which can be
simplified to x + 3y = 17.
Exercise 2:
dy
dy
(i) dx = 2x  3. When x = 4, y = 6 and dx = 5. The equation of the tangent is thus:
y  6 = 5(x  4) which can be simplified to 5x  y = 14.
dy
8
dy
1
(ii) dx =  x2 . When x = 4, y = 2 and dx =  2 . The equation of the tangent is thus:
1
y  2 =  2 (x  4) which can be simplified to x + 2y = 8.
dy
1
dy
1
(iii) dx =
. When x = 4, y = 2 and dx = 4 . The equation of the tangent is thus:
2 x
1
y  2 = 4 (x  4) which can be simplified to x  4y + 4 = 0.
dy
1
dy
1
(iv) dx =  2 (x 3)3/2. When x = 4, y = 1 and dx =  2 . The equation of the tangent is thus:
1
y  1 =  2 (x  4) which can be simplified to x + 2y = 6.
Exercise 3:
dy
dy
1
(i) dx = x3. When x = 2, y = 5 and dx = 8. The slope of the normal is thus  8 and so the
1
equation of the normal is y  5 =  8 (x  2) which can be simplified to x + 8y = 42.
dy
1
dy 1
(ii) dx =
. When x = 2, y = 2 and dx = 4 . The slope of the normal is thus  4 and so
2 x+2
the equation of the normal is y  2 =  4(x  2) which can be simplified to x + y = 10.
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dy
1
x + 2(x  1)
3x  2
dy
=x.
+ 1. x  1 =
=
. When x = 2, y = 2 and
= 2.
dx
dx
2 x1
2 x1
2 x1
1
1
The slope of the normal is thus  2 and so the equation of the normal is y  2 =  2 (x  2)
which can be simplified to x + 2y = 6.
(iii)
Exercise 4:
Let the point of contact of the tangent be Q(t, t2  5t + 7).
dy
dx = 2x  5 so the slope of the tangent at Q is 2t  5.
The equation of the tangent at Q is thus y  (t2  5t + 7) = (2t  5)(x  t).
Since this tangent passes through (2, 1) we have:
1  (t2  5t + 7) = (2t  5)(2  t) which can be simplified to t2  4t + 4 = 0.
This gives (t  2)2 = 0 so t = 2 is the only solution. There is therefore only one tangent through
(2, 1).
Substituting t = 2 into the equation of the tangent we get y  1 =  (x  2) which can be
simplified to x + y = 3.
Exercise 5:
1
Let the point of contact of the tangent be Q(t, t ).
dy
1
1
dx =  t2 so the slope of the tangent at Q is  t2 .
1
1
The equation of the tangent at Q is thus y  t =  t2 (x  t).
Since this tangent passes through (2,  4) we have:
1
1
 4  t =  t2 (2  t) which can be simplified to 4t2 + 2t  2 = 0. Dividing by 2 gives
2t2 + t  1 = (2t  1)(t + 1) = 0. This has two solutions: t = ½ and t =  1.
1
Substituting t = 2 into the equation of the tangent we get y  2 = 4(x  ½) which can be
simplified to 4x + y = 4.
Exercise 6:
Let the point of contact of the tangent be Q(t, t ).
dy
1
1
dx =  2 t so the slope of the tangent at Q is  2 t .
The equation of the tangent at Q is thus y 
t =
1
2 t
(x  t).
Since this tangent passes through (1, 1) we have:
1
1 t =
(1  t) which can be simplified to 3t  2 t  1 = 0.
2 t
Substituting T = t gives us 3T2  2T  1 = (3T + 1)(T  1) = 0.
1
This has two solutions: T = 1 and T =  3 . But since T = t can’t be negative, T = 1 is the
only solution. It corresponds to t = 1.
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Substituting t = 1 into the equation of the tangent we get y  1 = 
1
(x  1) which can be
2
simplified to x + 2y = 3.
Exercise 7:
dy
(a) dx = 6x2 – 30x + 23
dy
(b) If dx = –1 then 6x2  30 x + 24 = 0. Dividing by 6 gives x2  5x + 4 = (x  1)(x  4) = 0
so x = 1 or 4. When x = 1, y = 0 and when x = 4, y =  30. So the required points are (1, 0)
and (4,  30).
(c) The tangent at (1, 0) is x + y –1 = 0 and the tangent at (4, 30) is x + y + 26 = 0.
(d)
1
x + y –1 = 0
26
x + y + 26 = 0
Consider the dotted line passing through (0,1) that cuts both lines at right angles. Now
the shaded right- angled triangle has a hypotenuse with the other two sides equal. Suppose the
lengths of these other two sides is x. Using Pythagoras theorem we have x2 + x2 = 25 which
25
5 2
5
gives x =
=
=
. This is the distance between these two parallel tangents.
2
2
2
Exercise 8:
Let the normal be at the point Q(t, t2).
dy
1
=
2t
so
the
slope
of
the
normal
at
Q
is

dx
2t .
1
The equation of the normal at Q is thus y  t2 =  2t (x  t).
Since this normal passes through (0, 3/2) we have:
1
3/2  t2 =  2t (0  t) which can be simplified to 2t3 = 2t.
This has solutions t = 0 and  1. (Don’t forget the t = 0 possibility. If you divided by t you
were ignoring this case.)
So there are three normals that pass though (0, 3/2). The normal for t = 0 is the y-axis, x = 0.
For t = 1 the normal is y  1 =  ½ (x  1) which can be simplified to x + 2y = 3.
For t = 1 the normal is y  1 = ½ (x + 1) which can be simplified to x  2y = 3.
Exercise 9:
The house is at the point (1, 0) so we want to find the tangents that pass through this point.
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Let the point of contact of the tangent (the position of the train) be T(t, t2).
dy
dx = 2x so the slope of the tangent at T is 2t.
The equation of the tangent at T is thus y  t2 = 2t (x  t).
Since this tangent passes through (1, 0) we have:
0  t2 = 2t (1  t) which can be simplified to t2  2t = 0.
The solutions for t are t = 0 or 2.
If t = 0 the train is at (0, 0), the station.
Although the tangent at this point passes through (1, 0) when the train is at the station the light
might come through the side window of the house, but not the front window. So we reject this
case.
This leaves t = 2, which puts the train at (2, 4). The distance between the house at H(1, 0) and
the train at T(2, 4) is (2  1)2 + (4  0)2 = 17  4.123105626. Because of the scale this
represents 4.123105626 kilometres. Now it would be ridiculous to give the answer to such a
question to this degree of accuracy. After all the train and the house are rather large objects.
Probably an answer to the nearest 10 metres is appropriate. So we give our answer as 4120
metres.
Question 10:
The cross-section of the mound is described by the equation y = 9  x2 where the origin
is at the centre of the base of the mound. Distances are in metres.
The curve y = 9  x2 cuts the x-axis at x = 3 so the base of the mound is a circle with radius 3.
The top of the mound is at (0, 9) and the light is at (0, 12).
The furthest points of the shadow are where the tangent at some point cut the x-axis.
We don’t know where this point is, but we do know that the tangent passes through (0, 12). So
we want to find the equations of the tangents that pass through (0,12).
Let the point of contact of the tangent be T(t, 9  t2).
dy
dx = 2x so the slope of the tangent at T is 2t.
The equation of the tangent at T is thus y  (9  t2) = 2t (x  t).
Since this tangent passes through (0, 12) we have:
12  (9 t2) = 2t (0  t) which can be simplified to t2 = 3. The solutions for t are t =  3 .
If t = 3 the equation of the tangent is y  (9  3) = 2 3 (x  3 ) which can be
simplified to 2 3 x + y = 12.
12
This cuts the x-axis (y = 0) at x =
. Multiplying top and bottom by 3 gives
2 3
x = 2 3  3.464101615. This represents a distance of 3.46 metres from the centre of the base
of the mound (any greater accuracy wouldn’t be appropriate). But the mound has a radius of 3
metres so the furthest the shadow reaches from the base of the mound is 0.46 metres, which is
better expressed as 46 centimetres.
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