Find x-Coordinates of Points on y=3x/(2x-3) with Normal Parallel to 9y=4x+3

AI Thread Summary
To find the x-coordinates of points on the curve y=3x/(2x-3) where the normal is parallel to the line 9y=4x+3, first determine the slope of the given line, which is 4/9. The slope of the normal line is then -9/4. The next step involves finding the derivative of the curve and setting it equal to -9/4 to locate the corresponding x-values. This process includes calculating the derivative and applying the condition of parallelism. The solution hinges on these calculations to identify the required x-coordinates.
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Find the x-coordinates of the points on the curve y=\frac{3x}{2x-3} where the normal is parallel to the line 9y=4x+3.

Could someone tell me what to do?
 
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Ask Omicron,apparently he's found a way to solve the same problem.

Daniel.

P.S.My advice would normally include find the derivative,the equation of the normal and then put the condition of parallelism and find the "x" intercept.
 
Yes, apparently I found a way.
Find the derivitive
Gradient of 9y=4x+3 and normal => 4/9
Gradient of the tangent line => -9/4
Set -9/4 = derivitive

You know, its too coincidental that you got the same question as me? :rolleyes:
 
Last edited:
1. What is the slope of the line 9y= 4x+ 3?

2. What is the slope of a line normal to that?

3. At what points is the derivative of y(x) equal to the answer you got to 2?
 

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