Points of intersection of curves given in parametric equations

Question

This is the question,

The points of intersection of the curves whose parametric equations are $x=t^2+1, y=2t$ and $x=2s, y=2/s$ is given by:

(a) $(1, –3)$, (b) $(2, 2)$, (c) $(–2, 4)$, (d) $(1, 2)

The answer is $(2,2)$ as per this website

What I did is I solved the first set of equations and eliminated $t$ and the, equation came out to be, $x^2 = y^2/4 +1$ but the point $(2,2)$ doesn't seem to be satisfying it.

Doesn't it satisfy the equation as it should lie on it also along with the result of the second equation?

Answer 1

Consider the parametrics equation $$E_1:\begin{cases}x=t^2+1,\\ y=2t\end{cases},\quad E_2:\begin{cases}x=2s,\\y=2s^{-1}\end{cases}, \quad (t,s)\in {\bf R}\times {\bf R}^*$$ The intersection between the curves $E_1\cap E_2$ can be find solving the non linear system $$t^2+1=2s,\quad 2t=2s^{-1}$$ Since $2t=2s^{-1}$ then $t=s^{-1}$ and the substitution in $t^2 +1=2s$ give the equivalent equation $(s^{-1})^2+1=2s$ that can be written $\frac{1+s^2}{s^2}=2s$ or $1+s^2=2s^3$.

In order to solve $1+s^2=2s^3$ over ${\bf R}^{*}$ we can write the equation as $2s^3-s^2-1=0$ and the factorization show that $(s-1)(2s^2+s+1)=0$ and it holds if and only if $s-1=0$ or $2s^2+s+1=0$.

The equation $2s^2+s+1=0$ has not solution over ${\bf R}^{*}$ since the $(1)^2-4(2)(1)<0$ and it is a quadratic equation. Therefore, the only solution is $s=1$.

Since $t=s^{-1}$ and $s=1$ we can the only solution $(s,t)=(1,1)$. Therefore, the curves have intersection only in the point $(x,y)=(2,2)$.

Answer 2

EDIT!:

First blue curve eliminating parameter $t$ $$ x=t^2+1, y=2 t,~ y_1= 2 \sqrt{x-1} \tag1 $$

Second purple curve eliminating parameter $s$ $$ x=2 s, y=2/s,~ y_2= 4 /x\tag2 $$

From either of 1) or 2)

$$ y_1=y_2~~; 2 \sqrt{x-1}=4/x$$

$$\to (x_{intxn}=2, y_{intxn}= 2) $$