3.4. Tangent to cubic
Solutions
1. At the right is the graph of the parabola
5
y = (x–1)(4–x)
together with a line which is tangent to the curve at the origin.
(a) Use the above strategy to find the slope of the line.
(b) Is there another line through the origin which is tangent
to the curve? If so, what is its slope?
0
-1
0
1
2
3
4
-5
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5
2
2. At the right is the graph of the cubic equation
y = x(1–x)(2–x)
1
together with two lines through the origin, one of which is
tangent to the curve at the origin and the other of which is
tangent to the curve at another point. Use the above strategy
to find the slope of both lines.
[Answer: m = 2 and m = –1/4]
0
-1
0
1
2
3
-1
The strategy
Consider the family {y = mx} of all lines passing through the
origin. Let the two tangent lines have slope m+ and m–.
They seem to each intersect the curve exactly twice. Our
intuition from the picture tells us that these are the only two
lines with 2 intersections. Indeed the remaining lines in the
family appear to fall into three classes according to their
slope mi:
m+ < m1 < ∞
m– < m2 < m1
–∞ < m3 < m–
intersect the curve 3 times
intersect the curve 3 times
intersect the curve once
The algebraic analysis
To find the intersections between the curve and the line:
yC = x(1–x)(2–x)
yL = mx
we equate the y-values:
x(1–x)(2–x) = mx.
Rewrite:
x[(1–x)(2–x) – m] = 0
x[x2 – 3x + 2–m] = 0
The roots of this are
3  9  4(2  m) 3  1  4m

x = 0 and x 
2
2
We want to find the values of m for which there will be exactly two roots. That will happen when the discriminant is zero:
1+4m = 0
And that gives us m = –1/4. Okay: that must be the negative
slope m–. But how do we get m+?
Well there’s another way to get two roots out of the above
equation, and that is if one of the roots from the quadratic
formula is 0. For that to happen, we’ll need to use the minus
sign and we get
3  1  4m  0
3  1  4m
9 = 1 + 4m
m = 2.
This will be m+.
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3. Draw the circle with centre on the x-axis at the point (2,0)
and radius 1. Now draw a line through the origin tangent to
the circle. The problem is to find the slope of the line.
(a) Solve the problem using the methods of Euclidean geometry.
(b) Solve the problem using the approach of this section.
Find the equation of the circle and of the family of lines
through the origin, and then ask for the condition that there be
one intersection.
4. In terms of the parameter m, how many solutions does the
following equation have?
x(6–x) = m
This is exactly the equation we needed to solve in the Example, and we used a discriminant analysis. But isn't there a
simpler way? Draw the graph of the parabola
y = x(6–x) .
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