Why is the graph of a linear relationship curved?

Question

In light phenomena (lenses and mirrors), we know the formula $frac{1}{v} + frac{1}{u} = frac{1}{f}$ which describes the relation where,
$v$ = the distance of the image from the mirror
$u$ = the distance of the object from the mirror
$f$ = focal point of curved mirror
Assuming a concave mirror for simplicity, when we plot the $frac{1}{v}$ on the x-axis and $frac{1}{u}$ values on the y-axis, we get a linear curve (i.e a simple straight line).
However, when we plot the ${v}$ vs. $u$ graph (i.e $v$ and $u$ on y and x-axis respectively) why do we get a curve, when the relationship according to the formula is linear??

A curve would indicate quadratic relationship, but I can easily prove the the relationship is definitely of power 1. How and why does this happen?

Does anyone have any ideas?

robphy · Accepted Answer

Solving $displaystylefrac{1}{x}+frac{1}{y}=frac{1}{f}$ for $y$,
we have
$displaystyle y=-frac{fx}{left(f-xright)}$.
But that expression for $y$ is not a straight line:
since one cannot find constants $B$ and $C$ that
 will make $y=Bx+C$

agree with $displaystyle y=-frac{fx}{left(f-xright)}$ for all $(x,y)$,
as seen using https://www.desmos.com/calculator ,

In fact, $displaystylefrac{1}{x}+frac{1}{y}=frac{1}{f}$
and $displaystyle y=-frac{fx}{left(f-xright)}$
describe a hyperbola,

which can be seen if we rewrite the relation as
$$left(x-fright)left(y-fright)=f^{2},$$
which is akin to Newton's formula.
[Copy the above equations in Desmos, then enter this one: $left(x-fright)left(y-fright)=f^{2}$.]

user256872 · Answer

The relationship between $u$ and $v$ is not linear. A linear relationship would be something like $au+bv=c  {rm for}  a,b,cinmathcal R$. But $displaystyle frac{1}{v} + frac{1}{u} = frac{1}{f}$ is not a linear relationship for $v,u$.
However, a linear relationship is obtained between $1/v$ and $1/u$, but not between $u$ and $v$.

Why is the graph of a linear relationship curved?

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