Math CentralQuandaries & Queries


Question from Suez, a student:

Find a parabola that passes through the point (1,4) and whose tangent lines at x= -1 and x= -5 have slopes 6 and -2 respectively.

Hi Suez.

The equation of a parabola is in the form:
y - A = B ( x - C )2

where A, B and C are undetermined constants.

So when you take the derivative of both sides with respect to x by using the chain rule, you get:

d/dx ( y - A) = d/dx ( B ( x - C )2 )

dy/dx = B d/dx ( x - C )2

dy/dx = B (2)(x - C) d/dx (x - C)

dy/dx = 2Bx - 2BC

This, as we expected, is a linear equation (no squared terms) that describes the tangent line to the parabola. You should know that the slope of the parabola (the tangent line) is simply dy/dx, so you can
substitute in your two values:

6 = 2B(-1) - 2BC
-2 = 2B(-5) - 2BC

Now there are two unknowns in two equations, which you can solve in any of the usual ways. Once you know B and C, you can substitute them into the original expression along with the co-ordinates of the point
it passes through. That lets you find A, the last missing piece of the parabolic equation.

Hope this helps,
Stephen La Rocque.

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