Expert: Paul Klarreich - 3/2/2007

Question
what is the equation of both lines that pass through the point (2,-3) and are tangent to the parabola y = x^2 + x

Answer
Questioner:   Joel
Category:  Calculus
 
Subject:  equations of tangents
Question:  what is the equation of both lines that pass through the point (2,-3) and are tangent to the parabola y = x^2 + x
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Hi, Joel,

Let a be the x-coordinate of a point on the parabola such that the tangent goes through (2,-3).  Then, at x=a:

y = a^2 + a

The slope of the tangent is  m = 2a + 1, from dy/dx = 2x + 1.

The point-slope form of the equation of a line is:

y - y0 = m(x - x0)

Taking  (x0,y0) as  (a,a^2 + a) we get:

y - a^2 - a = (2a + 1)(x - a), and in slope-intercept form:

y - a^2 - a = (2a + 1)x  - 2a^2 - a

y - a^2     = (2a + 1)x  - 2a^2

y           = (2a + 1)x  - a^2   << First equation.

Taking  x0,y0 as  (2.-3) we get:

y + 3 = (2a + 1)(x - 2)

y + 3 = (2a + 1)x - 4a - 2

y     = (2a + 1)x - 4a - 5     << Second equation


But these must be the SAME equation, so they must have the same slope and the same intercept.  That means:

-a^2 = - 4a - 5

0 = a^2 - 4a - 5

(a - 5)(a + 1) = 0

a = 5, a = -1.

I think you can handle the rest.