Question 1039424
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If the foot of the perpendicular to the tangent through the focus is indeed the point *[tex \Large \left(\frac{t}{2},\,0\right)], then the zero of the function representing the tangent (i.e. the answer to part a) and the zero of the function representing the perpendicular to the tangent through the focus (i.e. the function given in part b) must both be equal to *[tex \Large \frac{t}{2}].  I'll let you do the algebra to verify that the zeros are both *[tex \Large \frac{t}{2}].


I can't be sure about part d because you don't specify point P.  However, I'll go out on a limb and assume it is the focus.  In that case, just use the midpoint formulae with the two points *[tex \Large \left(0,\,\frac{1}{4}\right)] and *[tex \Large \left(\frac{t}{2},\,0\right)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ =\ \frac{x_1\ +\ x_2}{2}\ =\ \frac{0\ +\ \frac{t}{2}}{2}\ =\ \frac{t}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ =\ \frac{y_1\ +\ y_2}{2}\ =\ \frac{\frac{1}{4}\ +\ 0}{2}\ =\ \frac{1}{8}]


So for any given value of *[tex \Large t], M is the point *[tex \Large \left(\frac{t}{4},\,\frac{1}{8}\right)].  Hence, for all real values of *[tex \Large t] the locus of points *[tex \Large M_t] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{1}{8}]


Assuming, of course, that P represents the focus.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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