f'(a) = 2a + 1
And the slope of the line between the point (a, a^2 + a) and (2, -3) is (a^2 + a + 3)/(2 - a) = 2a + 1
Can you solve that?
Hey,
I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."
I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.
Thanks for your help!
Let's say you have a point (p,r) on your parabola. (So, r is given by )
The slope, m, of the line tangent to the parabola when x = p is . (this from your derivative)
The slope, m, of the line passing through points (p, r) and (2, -3) is given by
So that
Plug the in the earlier results for r & m into this. Solve for p. (There will likely be two answers.) You then have the x value(s) for the intersection of the tanget line, with the parabola.
Another way to do a problem like this is to use Fermat's "method of ad-equations" which predates Calculus. Any line through (2, -3) can be written y= m(x-2)- 3= mx- 2m- 3. Of course, at a point where such a line crosses the graph of the two functions must give the same y-value: or . That's a quadratic equation with solution given by the quadratic formula: . But if it is tangent that x must be a double root- the only root of this equation. That means that the discriminant must be 0: . Solve that equation for m and then .
A quadratic equation may have, of course, two roots so there may be two tangent lines that pass through (2, -3).