Derivation of graph of v against u

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SUMMARY

The discussion focuses on deriving the graph of v against u, specifically finding the intersection point (2f, 2f) for the equations u = v and 1/u + 1/v = 1/f. Participants suggest inserting the first equation into the second to solve for v. A key insight is transforming the equation into the form v - A = B/(u - C) to facilitate graphing. The user expresses initial confusion but later indicates understanding after reviewing an old textbook.

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  • Knowledge of intersection points in coordinate geometry
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Homework Statement


This is the graph of v against u .
i am having problem of getting the shape of graph and getting the point (2f, 2f) . can anyone teach me how to derive it?


Homework Equations





The Attempt at a Solution


sorry. i don't know how to use LATEX. it may be quite untidy if i type it using / sign for all the equation.

so , i have did it beside the graph in the photo . hopefully you guys can understand .
 

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The question is a bit unclear. Is this the full problem exactly as stated?

It seems you want to find the point where the curves u=v and 1/u + 1/v = 1/f intersect. What happens if you insert the first relation into the second and solve for v?
 
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Orodruin said:
The question is a bit unclear. Is this the full problem exactly as stated?

It seems you want to find the point where the curves u=v and 1/u + 1/v = 1/f intersect. What happens if you insert the first relation into the second and solve for v?

this is not a question actually. it's just a note. Can you explain further? i am actually finding v against u in my working. to get the shape of the graph. unfortunealy, i got stucked.
 
desmond iking said:
this is not a question actually. it's just a note. Can you explain further? i am actually finding v against u in my working. to get the shape of the graph. unfortunealy, i got stucked.

See if you can get the equation into the form ##v - A = \frac B{u-C}##
 
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thanks , after referring to the old book, i managed to understand it.
 

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