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

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SUMMARY

The discussion focuses on finding the x-coordinates of points on the curve defined by the equation y = 3x/(2x-3) where the normal line is parallel to the line represented by 9y = 4x + 3. The gradient of the line 9y = 4x + 3 is determined to be 4/9, making the slope of the normal line -9/4. Participants suggest calculating the derivative of the curve, setting it equal to -9/4, and solving for the x-coordinates where this condition holds true.

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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 derivative
Gradient of 9y=4x+3 and normal => 4/9
Gradient of the tangent line => -9/4
Set -9/4 = derivative

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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