Proving a line is tangent to a curve

Question

1) Imagine a point T(g, f) on a Cartesian plane, where g, f ≠ 0. Deduce, in terms of g and f, the equation of the line, in which its x-intercept and y-intercept have midpoint T, which means that when you connect the x-intercept and y-intercept, the midpoint of that line will be (g, f).

2) For m≠0, imagine the function $h: ℝ \ {0} → ℝ$ given by $h(x) = m/x$. Prove that if T is a point on the curve y=h(x), it means that the line calculated in 1) is a tangent to this curve.

I think I've solved 1) as the equation I ended up with is:

$$ y= \frac {−f}{g}⋅x+2f$$

The problem suggests having knowledge of or using the following concepts to prove 2): products of sets/subsets, relations, monotonic functions, continuous functions, injective functions, surjective functions, bijective functions, and the intermediate value theorem. Any explanations as to how and why any one of these ideas is helpful towards solving 2) and any other help would be extremely appreciated. Also, the proof does not need to include any of the aforementioned concepts (although, it would be helpful) so if you have any other ideas, please let me know :)

Answer 1

If $T$ is on the curve, substitute the coordinates of $T$ in $y=m/x$ to obtain $m$.

Then, with this value os $m$, calculate the derivative of $h$ at $g$ and check that it is the same as the slope of the line obtained in (1).

Answer 2

Your answer for part 1) is good.

For part 2), there are several possible approaches. One approach would be to prove that the graph of $h$ is convex on the domain $(0,\infty),$ then show that if you choose a point on the graph, no matter how you choose it, the line through that point according to part 1) is always below the graph of $h$ except at the chosen point.

A less laborious solution (provided that you already know about derivatives and tangents to curves and you know how to take the derivative of the function $h(x)=\frac mx$) is to suppose that $(g,f)$ are the coordinates of a point on the graph of $h,$ find the derivative $h'(x)$ when $x=g,$ find the equation of the line through $(g,f)$ with slope $h'(g)$ (which is the tangent line), and show that it is the same as the line found in part 1).

Another approach is via affine transformations of the plane, which preserve lines, preserve midpoints, and preserve tangency. Consider that you have some point $(x,y) = (g,f)$ on the graph of $y = \frac mx.$ Note that this implies $m = fg.$ Transform the entire plane by the transformation $(x,y) \to \left(\frac xg, \frac yf\right).$ Show that this transforms the graph of $y = \frac mx$ to the graph of $y = \frac 1x$ and transforms the point $(f,g)$ to $(1,1).$ By symmetry, the line segment from $(0,2)$ to $(2,0),$ with midpoint at $(1,1),$ is tangent to the graph of $y = \frac 1x$ at its midpoint. Invert the transformation, and you take that line segment to the segment between $(0,2f)$ and $(2g,0)$ and you take the graph of $y = \frac 1x$ to the graph of $y = \frac mx,$ which (due to this being an affine transformation) is still tangent to the line segment at the segment's midpoint.

The solution using the discriminant of a quadratic equation (shown in another answer) is a nice approach.

Answer 3

Equation of given curve is $$y=\frac{m}{x}$$ The equation of given line is $$y=2f-\frac{x\times f}{g}$$ Solving these equations, we get, $$\frac{m}{x}=2f-\frac{x\times f}{g}$$ $$\Longrightarrow m\times g=2f\times g\times x-x^2\times f$$ $$\Longrightarrow fx^2-2fgx+mg=0$$ Discriminant of this quadratic equation is, $$D=4f^2g^2-4fmg$$ $$=4fg(fg-m)$$ But, the point $T$ lies on the given curve. Therefore, $fg=m$ and discriminant is $0$.

Thus, this equation has two coincident roots. Hence, the curve touches the line i.e. the line is a tangent to the curve.

Hope it helps:)