# Homework Help: Another Turning point question

1. Feb 2, 2013

### lionely

Find the equation of the tangent to the curve xy=4 at the point P whose coordinates are (2t,2/t). If O is the origin and the tangent at P meets the x-axis at A and the y axis at B, prove
(a) that P is the midpoint of AB
(b) that the area of Triangle OAB is the same for all positions of P.
Find the the equations of the normals to the curve xy=4 which are parallel to the line
4x-y-2=0.

I got the equation of the of the tangent at P to be yt2 - 4t -x = 0

Not sure how to do a and b I don't really have much information about AB other than it's points like on P basically.

2. Feb 2, 2013

### Staff: Mentor

The y-coordinate of A is 0. It lies on the tangent, so it satisfies yt2 - 4t -x = 0. What is its x-coordinate?

3. Feb 2, 2013

x is -4t

4. Feb 2, 2013

### SammyS

Staff Emeritus
Now that you have the coordinates for point A, find the coordinates for point B. (The x coordinate for point B is 0.)

5. Feb 2, 2013

### lionely

B is (0, 4/t) oh and when you work it out the mid point is in fact P... to do part b now do I find the length of the lines and work out the area?

6. Feb 2, 2013

### SammyS

Staff Emeritus
Draw a sketch .

The result should be pretty obvious.

7. Feb 2, 2013

### lionely

So no calculations are needed to prove it?

How would I do the final part if I don't know any points for the normals?

8. Feb 2, 2013

### SammyS

Staff Emeritus
I just noticed that must be an error in your equation for the tangent line. Assuming that t is positive, both intercepts should be positive.

You have the wrong sign on x.

9. Feb 2, 2013

### lionely

This should be okay now right?

10. Feb 2, 2013

### rollingstein

Area = 1/2 * 4t * 4/t = 8

Hence constant.

11. Feb 2, 2013

### rollingstein

yt2 - 4t + x = 0

12. Feb 2, 2013

### lionely

Yeah now I see it, thanks. But the final part of the question
Find the the equations of the normals to the curve xy=4 which are parallel to the line
4x-y-2=0.

How would I do this without having any points for the normals?

13. Feb 2, 2013

### rollingstein

I get the normal as y=4x-15

14. Feb 2, 2013

### lionely

How did you calculate it I don't understand we don't have any points.

:S

15. Feb 2, 2013

### rollingstein

I'm not sure I am right.

But, Slope * Slope_normal = -1

You know Slope of curve so find slope of normal.

Then equate it to slope of given line to find where the normal lies.

16. Feb 2, 2013

### lionely

slope of curve = -4x^2 = equation of given line 4x-2

solve for x? then get y from equation of the curve then find equation of normal?

IGNORE this makes no sense.

Last edited: Feb 2, 2013