Derive a formula for a function F

Question

Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$.

Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$.

Derive a formula for $F$.

Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.

How to derive a formula for $F$? If I get the line passing through $p$ and $(t,0)$, I get $(D): y = at-1$, right?

And I know that to show that $\phi$ : $\mathbb R$ → $S$ \ {$p$}, I have to show that $\phi$ is injective and surjective.

Please I need help solving this

Answer 1

For a start: From the intercept theorem we have $$\frac y1=\frac{\sqrt{(x-t)^2+y^2}}{\sqrt{t^2+1}}.$$ Squaring and using $x^2+y^2=1$ we arrive in $$y^2=\frac{(t-1)^2}{1+t^2}.$$

Answer 2

If we express $S$ in polar coordinates:

$$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$

then, from simple trigonometric relations, the inverse function $\Phi$ is:

$$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$

and its inverse $F$ is:

$$F(t)=\frac\pi{2}-2\arctan t$$

After changing coordinates back to cartesian, $F$ will look like:

$$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$

which can be further simplified to:

$$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$