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A parabola is given by the Cartesian equation: $y^2=16x$

So that can be written as $y = \pm\sqrt{16x}$

So the derivative (to find the gradient of the tangent at a point $x$) is

$$\dfrac{dy}{dx} = \pm 2x^{-0.5}.$$

So when putting this (as $m$) in the equation $y - y_1 = m(x - x_1)$

Should I use the positive or negative derivative (which do I choose out of the plus or minus)?

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I'm pretty sure you've described a parabola, not a hyperbola. – MartianInvader May 31 '12 at 23:25
It depends on whether you're looking in the upper or lower half plane. Sketch the graph, and things should become clear. You have a parabola "opening to the right". – David Mitra May 31 '12 at 23:27
You could also use implicit differentiation. – Neal May 31 '12 at 23:32
Sorry, it was late :) – Jonathan. Jun 1 '12 at 6:30
up vote 6 down vote accepted

$y^2=16x$ describes a parabola opening to the right, not a hyperbola. The vertex is at the origin. So the tangents with positive slope will be the ones that lie above the $x$-axis (those corresponding to positive values of $y$), and the tangents with negative slope will be those that lie below the $x$-axis (negative values of $y$).

I note that you can do this without having to figure all of that out, and without having to "solve for $y$", by using implicit differentiation. From $$y^2 = 16x$$ we can take derivatives on both sides, using the Chain Rule for $y$ to get: $$\begin{align*} y^2 &= 16x\\ \frac{d}{dx} y^2 = \frac{d}{dx} 16x\\ 2yy' &= 16\\ y' &= \frac{16}{2y}\\ y' &= \frac{8}{y}. \end{align*}$$ So, provided we are not at the point $(0,0)$ (where the tangent is vertical), the slope of the tangent will be $\frac{8}{y}$ at the point $(x,y)$ (where $y^2 = 16x$). This will automatically take care of the "sign".

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Thanks, I haven't learn implicit differentiation or the chain rule,but the part about above and below the x axis is helpful (and obvious now you've said it :)), but ill find out more about the rest, it looks much better. – Jonathan. Jun 1 '12 at 6:34

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