Intersections of a Line and a Parabola: Solving for t

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Homework Statement


Consider the line L(t)=<1+5t, 3+t>. Then L intersects:
1. The x-axis at the point (-14,0) when t= -3
2. the y-axis at the point (0, 14/5) when t=(-1/5)
3. the parabola y=x^2 at the points ____ and ____ when t=____ and ____.


Homework Equations



y=x^2?

The Attempt at a Solution



I really have no idea what this is asking. I tried setting y = x^2 so (3+t) = (1+5t)^2 with the results of (110/50) and (272/50) but it didn't work. I really have no idea what it's asking. I'm sure it's a very simple problem because this is number 2/16 on my homework and I've already gotten all the others correct (including #1 and #2 of this problem), I just need a little push in the right direction.
Thanks!
 
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I think you are doing the right thing. But you are getting the wrong solutions for t. Unfortunately, when I work it out I get a pretty messy solution for the two values of t. Can you show how you solved it?
 
NEVER MINE I figured it out... basic algebra problem it works now!
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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