# Thread: Extremely Hard Max/Min Problem

1. ## Extremely Hard Max/Min Problem

5. A chord joins any two points A and B on the parabola whose equation is y^2 = 4x. If C is the midpoint of AB, and CD is drawn parallel to the x-axis to meet the parabola at D, prove that the tangent at D is parallel to chord AB.

I've attempted to graph the situation (attempted graph is attached), however I have no idea how to represent it with equations. What are my constraints? What maximum/minimum am I trying to solve for?

I've understood this unit up to now, though I simply can't find an answer to this question. Help would be strongly appreciated.

2. Originally Posted by RogueDemon
5. A chord joins any two points A and B on the parabola whose equation is y^2 = 4x. If C is the midpoint of AB, and CD is drawn parallel to the x-axis to meet the parabola at D, prove that the tangent at D is parallel to chord AB.

I've attempted to graph the situation (attempted graph is attached), however I have no idea how to represent it with equations. What are my constraints? What maximum/minimum am I trying to solve for?

I've understood this unit up to now, though I simply can't find an answer to this question. Help would be strongly appreciated.
Your really don't need to maximize or minimize anything!

First let $y_1$ be the y coordinate of the first point then this gives the ordered pair

$\displaystyle \left( \frac{y^2_1}{4},y_1\right)$

$y_2$ be the y coordinate of the and point then this gives the ordered pair

$\displaystyle \left( \frac{y^2_2}{4},y_2\right)$

So the mid point is

$\displaystyle \left( \frac{y_1^2+y^2_2}{8},\frac{y_1+y_2}{2}\right)$

and the slope is

$\displaystyle \left( \frac{y_2-y_1}{\frac{y^2_2-y_1^2}{4}}\right)=\frac{4}{y_1+y_2}$

Now the implicit derivative of the parabola is

$\displaystyle 2yy'=4 \iff y'=\frac{2}{y}$

but y at the midpoint is $\displaystyle \frac{y_1+y_2}{2}$

3. Thank you very much!

Your solution makes sense, though are you sure that max/mins aren't involved in this? I need to be sure so that I show all necessary steps on the upcoming test.

4. Originally Posted by TheEmptySet
Your really don't need to maximize or minimize anything!

First let $y_1$ be the y coordinate of the first point then this gives the ordered pair

$\displaystyle \left( \frac{y^2_1}{4},y_1\right)$

Typo.
Slope is 4/y1+y2, not 2/"".

$y_2$ be the y coordinate of the and point then this gives the ordered pair

$\displaystyle \left( \frac{y^2_2}{4},y_2\right)$

So the mid point is

$\displaystyle \left( \frac{y_1^2+y^2_2}{8},\frac{y_1+y_2}{2}\right)$

and the slope is

$\displaystyle \left( \frac{y_2-y_1}{\frac{y^2_2-y_1^2}{4}}\right)=\frac{2}{y_1+y_2}$

Now the implicit derivative of the parabola is

$\displaystyle 2yy'=4 \iff y'=\frac{2}{y}$

but y at the midpoint is $\displaystyle \frac{y_1+y_2}{2}$
There's a typo there i believe.
Slope is 4/(y1+y2).

5. yes you are correct. I will edit above

6. Originally Posted by RogueDemon
Thank you very much!

Your solution makes sense, though are you sure that max/mins aren't involved in this? I need to be sure so that I show all necessary steps on the upcoming test.
Yes I am sure. You were not asked to maximize or minimize anything. Just prove that the tangent line is parallel to a chord. I did use implicit differentiation but that was the only calculus in the whole problem