   SEARCH HOME Math Central Quandaries & 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

6 = 2B(-1) - 2BC
and
-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.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.