Proving Question!

[sadness]

Please correct me and guide me to the right answer!

Let C be a point on the positive x - axis and 0 be a point on the positive y-axis such that line CD is tangent to the curve y = 8/x, x> 0 at any point P. Prove that P is the midpoint of segment CD.

Midpoint formula : ( (x1 + x2) / 2, (y1 + y2) /2)
y' = -8 / x^2

let C be (c,0)
let D be (0,d)
let P be ( c/2 , d/2)

let use (c/2, d/2) to find the slope using the derivative
slope = y'(c/2) = -8 / (c/2)^2
= -32 / c^2

use slope formula, find the line of the tangent

y - d/2 = (-32 / c^2) ( x - c/2)
y = -32x / c^2 + 16/c + d/2

yint = 16 / c + d/2
= 32 + cd / 2c
x int = 0 = -32x / c^2 + 16/c + d/2
32x/c^2 = (32 + cd) / 2c
x = c(32 +cd) / 32

midpoint ( (32c + 32cd / 32)/2 , (32 + cd / 2c) / 2)
= 32c + 32cd / 16 , 32 + cd / c
The last equation does not equal to c/2, d/2.......
I don't know how to isolate it or simplify it.

Please correct me! Thank you!

 

[Unknown008]

There's something wrong here:

From:
\(\displaystyle \dfrac{32x}{c^2} = \dfrac{32 + cd}{2c}\)

To:
\(\displaystyle x = \dfrac{c(32 +cd)}{32}\)

It should be:
\(\displaystyle x = \dfrac{c(32 + cd)}{64}\)


I am a bit confused by your methods however. I would first get the slope of the line going through CD, in terms of c and d. Then find the derivative of the graph in terms of x, equate this to the slope of line CD to get an x coordinate. The x-coordinate should be equal to c/2. This way, you are showing that for any value of c and d, the curve y = 8/x will always be a tangent of line CD and touch each other at the midpoint of CD.

 

[sadness]

There's something wrong here:

From:
\(\displaystyle \dfrac{32x}{c^2} = \dfrac{32 + cd}{2c}\)

To:
\(\displaystyle x = \dfrac{c(32 +cd)}{32}\)

It should be:
\(\displaystyle x = \dfrac{c(32 + cd)}{64}\)


I am a bit confused by your methods however. I would first get the slope of the line going through CD, in terms of c and d. Then find the derivative of the graph in terms of x, equate this to the slope of line CD to get an x coordinate. The x-coordinate should be equal to c/2. This way, you are showing that for any value of c and d, the curve y = 8/x will always be a tangent of line CD and touch each other at the midpoint of CD.

I originally plugged it in with two points and tried to redo the steps using the variables which clearly is unsuccessful > <

From your method, how were you able to isolate d out to get the x coordinate? I used slope formula in terms of c and d to get the slope of d/-c ? Sorry for the need of clarification

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[Unknown008]

No, that's okay, we're here to learn ^^

1. Gradient of line in terms of c and d:
\(\displaystyle m = \dfrac{0-d}{c-0} = -\dfrac{d}{c}\)

2. I would get the gradient of the curve, which is \(\displaystyle y' = -\dfrac{8}{x^2}\)

3. Equate both: \(\displaystyle -\dfrac{d}{c} = -\dfrac{8}{x^2}\)

\(\displaystyle x = \sqrt{\dfrac{8c}{d}}\)

This is the x-coordinate where the curve and the line have the same gradient.

4. The y coordinate would be: \(\displaystyle y = 8\div\sqrt{\dfrac{8c}{d}}\)

\(\displaystyle y = \sqrt{\dfrac{8d}{c}}\)

This is the y-coordinate where the curve and the line have the same gradient.

Now the equation for the line CD would be: \(\displaystyle y = d - \dfrac{d}{c}x\)

And it meets the curve \(\displaystyle y = \frac{8}{x}\) at:

\(\displaystyle \dfrac{8}{x} = d - \dfrac{d}{c}x\)

\(\displaystyle \dfrac{d}{c}x^2 - dx + 8 = 0\)

At a tangent, \(\displaystyle b^2-4ac = 0\)

\(\displaystyle (-d)^2 - 4\left(\dfrac{d}{c}\right)(8) = 0\)

\(\displaystyle d = \dfrac{32}{c} \Leftrightarrow c = \dfrac{32}{d}\)

5. Taking the x and y coordinates:

\(\displaystyle x = \sqrt{\dfrac{8c}{\left(\dfrac{32}{c}\right)}}\)

\(\displaystyle x = \dfrac{c}{2}\)

\(\displaystyle y = \sqrt{\dfrac{8d}{\left(\dfrac{32}{d}\right)}}\)

\(\displaystyle y = \dfrac{d}{2}\)

The midpoint of \(\displaystyle (c, d)\) is indeed \(\displaystyle \left(\dfrac{c}{2}, \dfrac{d}{2}\right)\)

I hope this makes sense!