Expressing q in Terms of a_1,b_1,a_2,b_2

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Homework Statement


A tangent line at point A with coordinate (a,f(a)) of function f(x) intersects f(x) at point B coordinate (b,f(b)) . A vertical line is drawn from point p (a<p<b) and intersects f(x) at C. From C a perpendicular line to the tangent line is drawn which intersect the tangent line at point D with coordinate (q,f(q))

The tangent line can be described by y_1(x) =a_1 x+ b_1

|CD| can be described by y_2(x) =a_2 x+ b_2

Express q in terms of a_1,b_1,a_2,b_2

http://imageshack.us/photo/my-images/706/5cqg.jpg/

Homework Equations


-

The Attempt at a Solution


I can find a_1,b_1,a_2,b_2 in terms of a, b, f(a) and f(b).

I can express q in terms of the slope of the tangent:

q = p + H \sin(\alpha) \cos(\alpha)

with
\alpha = arctan\left( \frac{f(b)-f(a)}{b-a} \right)

H = f(p) -y_1(p)
and now what?
 
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dirk_mec1 said:

Homework Statement


A tangent line at point A with coordinate (a,f(a)) of function f(x) intersects f(x) at point B coordinate (b,f(b)) . A vertical line is drawn from point p (a<p<b) and intersects f(x) at C. From C a perpendicular line to the tangent line is drawn which intersect the tangent line at point D with coordinate (q,f(q))

The tangent line can be described by y_1(x) =a_1 x+ b_1

|CD| can be described by y_2(x) =a_2 x+ b_2

Express q in terms of a_1,b_1,a_2,b_2

http://imageshack.us/photo/my-images/706/5cqg.jpg/


Homework Equations


-


The Attempt at a Solution


I can find a_1,b_1,a_2,b_2 in terms of a, b, f(a) and f(b).

I can express q in terms of the slope of the tangent:

q = p + H \sin(\alpha) \cos(\alpha)

with
\alpha = arctan\left( \frac{f(b)-f(a)}{b-a} \right)

H = f(p) -y_1(p)
and now what?
This is a fun Euclidean geometry problem. :biggrin:

##a_1q+b_1=a_2q+b_2##. Solve for ##q##.
 
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I wonder if there is a mistake in this question. Expressing q in terms of ##a_1##, ##a_2##, ##a_3## and ##a_4## means we don't need to know what those numbers are.
 
Mandelbroth said:
This is a fun Euclidean geometry problem. :biggrin:

##a_1q+b_1=a_2q+b_2##. Solve for ##q##.

I can seriously hit myself against the wall. Thanks man.
 
dirk_mec1 said:
I can seriously hit myself against the wall. Thanks man.
No problem. They threw a lot of unnecessary information out there. It took me a minute too. :-p
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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