Why is it that the tangent lines to the hyperbola y=1/x at any arbitrary point (x0,y0) intersect the x-axis at (2x0,0) and the y-axis at (0,2y0)?

Question

I cannot understand how this type of symmetry arises, all I know is that it has to do with the hyperbola y=1/x and that it is a self-inverse function.

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Answer 1 · 2017-07-02T16:23:57

Refer to the Proof given in the Explanation Section.

Let, #P(x_0,y_0) in" Hyperbola "H={(x,y) : y=1/x} sub RR^2,#be

an arbitrary point (pt.). #:. y_0=1/x_0...........(ast).#

We will show the Tangent (tgt.) line #t# at the pt. #P# intersects the

Axes in the pts. #(2x_0,0) and(0,2y_0).#

We know that, #dy/dx# gives the slope of tgt. to the curve at the pt.

#(x,y).#

# y=1/x, :. dy/dx=-1/x^2, &, :., [dy/dx]_{(x_0,y_0)}=-1/x_0^2.#

Thus, the slope of tgt. at #P# is #-1/x_0^2, &, P in t.#

Therefore, the eqn. of tgt. #t# at #P# is,

# t : y-y_0=-1/x_0^2(x-x_0), i.e., because" of "(ast), #

# t : x_0(y-1/x_0)=-x/x_0+1," or, again by, "(ast)#

# t : 1/y_0(y-1/x_0)=-x/x_0+1.#

# t : y/y_0-1/(x_0y_0)=-x/x_0+1.#

# t : x/x_0+y/y_0=2, because, x_0y_0=1.#

# t : x/(2x_0)+y/(2y_0)=1,# showing that the Intercepts made by

#t# on the Axes are #2x_0 and 2y_0.#

Thus the tgt. line intersects the Axes in the pts.

#(2x_0,0) and (0,2y_0).#

This proves the Result.

Enjoy Maths.!