7. Determine the equation of a parabola, in standard form (y = ax2 + bx + c), such that the line y = 2x − 3
is tangent to the parabola at some point.
Solution 1:
Consider the parabola with equation y = x2 , then y 0 = 2x. Since we want y = 2x − 3 to be tangent to our
given parabola, the slope of the tangent we want is mtangent = 2. If we set y 0 = 2 we get 2x = 2 so x = 1.
That means the tangent to the point (1, 1) on the parabola has slope 2 so is parallel to y = 2x − 3 (but isn’t
the same line you could check this, if you found the equation of the tangent it would be y = 2x − 1).
Substitute x = 1 into the equation of the line to get y = −1, so (1, −1) is a point on the line just below the
point on the parabola whose tangent is parallel to the line. If we consider a new parabola, that is congruent
to y = x2 but moved down 2 units, then (1, −1) would be on this parabola and the tangent would have slope
2, so it must be the line y = 2x − 3. (Note also, if we determined the equation of the tangent line to y = x2 at
(1, 1), i.e., y = 2x − 1, this line is 2units above the line we want, so we could reason that we need to move it
down.
y
y = x2
y = 2x − 1
y = 2x − 3
(1, 1)
y = x2 − 2
x
(1, −1)
∴ The parabola with equation y = x2 − 2 is tangent to the line y = 2x − 3 (at the point (1, −1), see the figure
above).
Note: If we translate the graph of y = x2 so that the point (1, 1) ends up anywhere on the line, then we will
have a solution to the problem. For example, if we pick a random number . . . like x = −13, for example, then
substituting into the equation of the line we get y = −29, so (−13, −29) is a point on the line. If we translate
the parabola so that the image of the point (1, 1) is (−13, −29), then the image parabola will be a solution. In
this case to go from (1, 1) to (−13, −29), we go left 14 and down 30, so the new parabola would have equation
y = (x + 14)2 − 30, which you can check to see that it works.
Solution 2:
Let’s pick the point on the line where we want the parabola to be tangent (it can be anywhere). For
example, if we let x = 2 then y = 2(2) − 3 = 1, so (2, 1) is the point of tangency. We know our parabola must
pass through this point, so substituting x = 2 and y = 1 into y = ax2 + bx + c gives
4a + 2b + c = 1.
(1)
Differentiating y = ax2 + bx + c, gives y 0 = 2ax + b. We need the slope of the tangent at x = 2 to be 2 (since
y = 2x − 3 is the tangent line), so we must have
2 = 4a + b.
(2)
Looking at (2) gives
b = 2 − 4a
which, when substituted into (1) yields
c = 4a − 3.
Since we have only two conditions (the point must be on the parabola and the slope of the tangent must be 2
when x = 2), then there is a degree of freedom, that is, each value of a determines a new solution. For example
choosing a = 1, a = −1 and a = 21 yields
1
a= ,
2
b = 0,
a = 1,
a = −1,
b = −2,
b = 6,
c = 1,
c = −7,
y = x2 − 2x + 1,
y = −x2 + 6x − 7,
y = x2 − 2x + 1
y
c = −1,
1
y = x2 − 1.
2
y = 2x − 3
y = 21 x2 − 1
(2, 1)
x
y = −x2 + 6x − 7
Solution 3(only for the brave):
Extending the idea of solution 2, if we want the parabola to be tangent when x = k for some x ∈ R, then
the point (k, 2k − 3) must be on the parabola. Substituting into the equation of the parabola yields
k 2 a + kb + c = 2k − 3.
(3)
Also, we know the slope of the tangent at the point (k, 2k − 3) must be 2, so since the derivative is y = 2ax + b,
we must have
2ak + b = 2.
(4)
Rearranging (2) gives
b = 2 − 2ak
which when we put into (1) gives
c = k 2 a − 3.
Which means we have two degrees of freedom, we can pick where the point of tangency is (k) and the “shape”
of the parabola (a).
For example, if we pick k = 1, we are talking about the point (1, −1) (solution 1). So if we pick a = 1, like
our solutions from example 1 we get b = 2 − 2(1)(1) = 0 and c − (1)2 (1) − 3 = −2 and the solution is y = x2 − 2,
just as we saw in solution 1.
Similarly, if we pick k = 2, we are talking about point (2, 1) from solution 2. A quick check shows that
b = 2 − 4a and c = 4a − 3, which reproduces the solutions from solution 2. Thus the equation
y = ax2 + (2 − 2ak)x + (k 2 a − 3)
is the equation of a parabola that will be tangent to the line y = 2x − 3 at the point (k, 2k − 3) for each choice
of a and k.