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 45^{o} (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.