Extremely Difficult Calc Problem?

Question

I've tried alooot of answers for this one, and am almost out of submits. Help please!

Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x2 + x.
y = (smaller slope)
y = (larger slope)

hmm..i've already tried both -1, and 5. It does ask for the slope, so i dont know..?

lol well thank you both, that answer is right :)

Answer 1

Let (p, p^2+p) be the point(s) on the parabola in which we are interested.
So the gradient of the parabola at these points must be equal to the gradient of the line.

y = x^2 + x
dy/dx = 2x + 1
gradient of parabola = 2p+1

gradient of line = [p^2 + p + 3] / [p - 2]

So [p^2 + p + 3] / [p - 2] = 2p+1
p^2 + p + 3 = 2p^2 - 3p - 2
p^2 - 4p - 5 = 0
(p-5)(p+1) = 0

p = -1, 5

The respective slopes would be 2p+1
= -1 and 11

equations of lines:
1) y + 3 = -1*(x - 2)
y = -x - 1
2) y + 3 = 11*(x-2)
y = 11x - 25

Answers:
y = - x - 1
y = 11x - 25

*EDIT*
Yes scythian, I corrected the error. Ahem, MY 10 points please.

Answer 2

The equation of any line through (2,-3) is k x - 2 k - 3, where k is the slope. We subtract this from x^2 + x, and solve for roots:

x^2 + x - k x - 2 k - 3 = 0

The roots contain the quadratic √(k^2 -10 k - 11), which has to be zero in order for the line to be a tangent (double root). Thus, solving for k, we find that k = -1, 11, so that the line equations are:

y = - x - 1
y = 11 x - 25

(pss..., Dr. D made a mistake, give me the 10 points)

Addendum: Shawn, you wound me. This has been an extremely difficult problem. Why, I had to think really hard on this for a few seconds.

Answer 3

I agree with these guys. I think they're doing math problems which are much too easy for them, and I should get the 10 points for pointing it out.