Parabolas and tangent lines

ksdhart

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I am having difficulty with a problem from my Precalc II course. The problem's text is:

93) Show that an equation of the tangent line at the point (x1, y1) on the parabola x = 4ay^2 is y = (1 / 8ay1)x + y1 / 2

The only progress I've made on this problem is to transform the equation of the parabola into the standard form of y^2 = (1 / 4a)x. But beyond that, I don't even know where to start.
 
93) Show that an equation of the tangent line at the point (x1, y1) on the parabola x = 4ay^2 is y = (1 / 8ay1)x + y1 / 2
First, the problem breaks down into three parts
(1) We need the slope of the tangent line at x1 - that is the value of the derivative of the function at x1

(2) The formula for a line with a given slope and point through the which the line passes [in this case the point is (x1, y1) ].
and
(3) Turn the formula in (2) into the form given by the problem.


So, what is the derivative of x with respect to y [I guess I'm old fashioned because it sort of bugs me to see x as a function of y but I can live with it]? Well x' = 8 a y so
x'(y1)= 8 a y1
and we have the slope of the line we want.

Next, a standard form for a line with a given slope m and point through which the line passes can be written as
x = m (y - y1) + x1
where (y1 , x1) or, as I prefer, (x1 , y1) is the point through which the line passes. But we know that the slope (value of the derivative) is given by
m = 8 a y1
and that
x1= 4 a y12
so we can write the equation as
x = 8 a y1 (y - y1) + 4 a y12

Finally we need to put the equation in the form of the answer. FINALLY, someone got some sense and wants to write y as a function of x:D

I'll leave that part to you.
 
First off, thank you for your thorough answer. The only difficulty is that I only have a basic understanding of derivatives, and being that this is Precalc, I won't learn about them until next quarter. I feel like I'm being a huge bother... but, is there perhaps another way of solving the problem without using calculus?
 
Prior to Newton and Leibniz development of Calculus, Fermat developed a method for finding tangent lines of, specifically, parabolas. It was based on the fact that if we treat y= cx+ d and x= 4ay^2 as simultaneous equations, then they have either none, one, or two solutions. Of course, the "one solution" case is precisely when the line is tangent to the parabola.

Here, we can solve y= cx+ d and x= 4ay^2 by replacing x in the first equation by ay^2 to get y= acy^2+ d. That gives a quadratic equation, acy2y+d=0\displaystyle acy^2- y+ d= 0 in standard form, which, by the quadratic formula, has solution y=1±14acd2ac\displaystyle y= \frac{1\pm\sqrt{1- 4acd}}{2ac}. That will have two solutions (the line passes through the parabola) if 14acd>0\displaystyle 1- 4acd> 0, no solutions (the line "misses" the parabola) if 14acd<0\displaystyle 1- 4acd< 0, and one solution (the line is tangent to the parabola) if and only if 14acd=0\displaystyle 1- 4acd= 0. Since we are given a as part of the definition of the parabola, a line, y= cx+ d, will be tangent to the parabola, x= 4ay^2, if and only if cd=14a\displaystyle cd= \frac{1}{4a}.
 
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Thanks for the help, guys. The whole class worked the problem together this morning, and we ended up using the information to form a quadratic equation and then solving for the slope when the discriminant (b - 4ac) is equal to 0. And that seems to be exactly what HallsOfIvy suggested. I never would have came to that conclusion on my own, though.
 
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