Parametric Curves: Tangent Lines

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Homework Statement



Find equations of the tangents to the curve x=3t^2+1, y=2t^3+1 that pass through the point (4,3).

The Attempt at a Solution



I was able to find the equation y=x-1 as a tangent line through the point (4,3) for the part of the curve above the x-axis since (4,3) is on the curve.

However, I do not know how to find the equation of the tangent for the part of the curve below the x-axis since I do not know what point on the curve the tangent passes through and therefore do not know the t value.

Any help is appreciated, thanks
 
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Write down the equation for the tangent line to the curve at general value of t. Now put in the condition (4,3) is on the line. You should get a couple of cubic equations. You already know t=1 is a solution. Is there another?
 
Well, you get a cubic equation. Knowing that t= 1 is a solution makes it easy to solve. Dick's suggestion is right on the money.
 
Duh, right. Make that 'one' cubic.
 
thanks a lot for the help, i was able to find the second equation, thanks
 
So what *is* that second equation? I'm stuck!
 
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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