WORKSHOP 2·2
Solutions
Warm-up
Question: A street light is mounted at the top of a 15-ft-tall pole. You are standing 40 ft from
the pole. How long is your shadow?
Solution: Here is a figure of the situation:
15
h
l
40
Let h be the height of the student and let l be the distance between the student and the end of the shadow.
We then have two similar right triangles of height h and 15 and base l and l + 40. Since the ratios of lengths
must be equal we obtain:
l
l + 40
=
h
15
1
8
1
l( − ) =
h 15
3
8
15h
40h
l= ·
=
.
3 15 − h
15 − h
3. Part B [20 minutes]
Question: If (−3, 0) is the x-intercept of a line tangent to the parabola y = −x2 + 5, what is the slope of the
line? There may be more than one answer.
Solution:
We don’t know the point of intersection between the line and the parabola, let’s call it (a, b). We can use
the formula of the tangent line at (a, b)
y = m(x + 3)
with the slope given by the derivative of the parabola
m = −2a
and making it equal to the parabola to get the x value of the intersection, which we can then use to solve
for the slopes.
1
5
−3
−2
2
3
−a2 + 5 = −2a(a + 3)
−a2 + 5 = −2a2 − 6a
a2 + 6a + 5 = 0
(a + 5)(a + 1) = 0
a = −5, m = −2(−5) = 10
a = −1, m = −2(−1) = 2.
Note that there are in fact two slopes, the other one intersects outside of the picture.
2
Main Problem
Question: The figure shows a lamp located 3 units to the right of the y-axis and a shadow
created by an elliptical object. The surface of the object is described by points (x, y) whose
co-ordinates are related by the equation x2 + 4y 2 = 5. If the point (−5, 0) is on the edge of the
shadow, how high is the street lamp?
Solution: We observe that the the shadow is formed by the edge of the object. So the line that connects
the lamp and the point on the ground also touches the edge of the object. That is, it’s a tangent line!
Since we are dealing with the tangent line, we need two conditions to be true. First, the line must touch the
object at one point. Let’s call that point (a, b). Second, the slope of the object and the line must be the
same at point (a, b). Let’s call this slope m.
So start by finding the slope of the ellipse at a point (a, b), using implicit differentiation, and make it equal
to the slope of a line through the point (a, b) and (−5, 0):
dy
=0
dx
dy
−2x
m=
=
dx
8y
dy
−a
m=
=
.
dx
4b
2x + 8y
Alternatively, we know that the tangent line passes through the points (−5, 0) and (a, b), therefore we have
a formula for its slope m:
m=
b
(rise over run).
5+a
We can then make the two slopes together and use some algebra to obtain the following.
−a
b
=
4b
5+a
−a2 − 5a = 4b2
m=
2
2
a + 4b = −5a.
(1)
(2)
(3)
The point (a, b) is on the ellipse so we know that it follows the ellipse formula a2 + 4b2 = 5. The right side
of that formula is the same as in equation (3). Therefore we can make the left sides equal:
−5a = 5
a = −1.
3
We can use this value of a in the ellipse equation to get b:
a2 + 4b2 = 5
(−1)2 + 4b2 = 5
1 + 4b2 = 5
4b2 = 4
b2 = 1
b = ±1.
We find two values. These represent the two points on the ellipse whose tangent line cross the point (−5, 0).
There’s one going above the ellipse that creates the shadow and one below the ground, that cannot possibly
create a shadow, so we only keep the positive value of b = 1. We can then use any of the slope formula with
our a and b value to obtain m = 1/4.
Then you substitute x = 3 in the tangent line equation y = m(x + 5) to get the height of the lamp. it will
be 2 units.
An alternate solution is to use similar triangles. We have two right triangles that share the point (−5, 0).
One also has the points (a, b) and (a, 0) and the other one (3, 0) and (3, h), where h is the height of the lamp.
Using the previously found values of a and b we know that the smaller triangle has height of 1 and a base of
length 4. We also know that the larger triangle has a base of 8 and a height of h. Using ratios of lengths in
similar triangles we obtain the following formula:
h
1
=
4
8
h = 2.
4