Why the uv graph of concave mirror is parobilic and1/u and1/v graph is a linear graph???

The uv graph is actually a hyperbola

1/v + 1/u = 1/f

u + v = uv/f

uv = f(u+ v)

We know that equation of the type xy = a(x + y) is a hyperbola asymptotic to x = a, y = a

Therefore the equation

uv = f(u + v) is also a hyperbola asymptotic to u = f, v = f

Plot u on x axis and v on y axis you will get a curve of the shape shown in figure.

---------CONSIDER THE BRANCH OF HYPERBOLA IN THE FIRST QUADRANT-------

A line drawn at angle 45o (u = v) from the origin intersects the hyperbola at point C.

Find the intersection point of v = u and uv = f(u + v). Let the intersection point be C

u.u = f (u + u)

u = 2f

v = 2f

The coordinates of point C are (2f, 2f). The focal length of mirror can be calculated by measuring the coordinates of point C on your graph.

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SECOND METHOD (USING STRAIGHT LINE)

1/v + 1/u = 1/f

or

1/v = -1/u + 1/f

Comparing above equation with the equation of line y = mx + c

1/v is plotted on y axis

1/u is plotted on x axis

Slope of line m is -1

Constant term will be 1/f

Measure the constant term in your graph. The reciprocal of which will give the focal length.

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The horizontal asymptote is v = f (not u = f) and vertical is u = f (not v = f)

I just wrote them opposite.

Asymptotes are imaginary straight lines that are drawn to guide the shape of curve. The branch of curve tends to run parallel to the asymptote and touch each other at infinity. This is just for your extra knowledge.

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hmmm
 
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