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# The point of intersection of the curves whose parametric equations are $x = {t^2} + 1$, $y = 2t$ and $x = 2s$, $y = \dfrac{2}{s}$ is given byA.$\left( {4,1} \right)$B.$\left( {2,2} \right)$C.$\left( { - 2,4} \right)$D.$\left( {1,2} \right)$  Verified
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Hint: We will first find the value of $t$ from the given value of $x$, then will substitute the value in the equation of $y$ involving $t$. Similarly, find the value of $s$ from one equation and substitute in another to get an equation of $x$ and $y$. Solve the two equations to determine the values of $x$ and $y$.

First of all, we will find the value of $t$ from the given equation $y = 2t$
Then, $t = \dfrac{y}{2}$
Substitute the value of $t$ in the equation of $x$, that is $x = {t^2} + 1$
Therefore, now we have $x = {\left( {\dfrac{y}{2}} \right)^2} + 1$
On simplifying, we get,
$x = \dfrac{{{y^2}}}{4} + 1$
$x = \dfrac{{{y^2} + 4}}{4}$ (1)
Similarly, we will find the value of $s$ from the equation $y = \dfrac{2}{s}$
Now, we will get $s = \dfrac{2}{y}$
Next, we will substitute the value of $s = \dfrac{2}{y}$ in the equation $x = 2s$ to find the value of $x$.
Then, we have
$x = 2\left( {\dfrac{2}{y}} \right)$
$\Rightarrow x = \dfrac{4}{y}$
$\Rightarrow xy = 4$ (2)
Now, we will equate equations (1) and (2) we will get,
$\left( {\dfrac{{{y^2} + 4}}{4}} \right)y = 4$
Hence, on solving the equation we get,
$y\left( {{y^2} + 4} \right) = 16$
Now, we add and subtract $4y$ in the bracket to complete the square and hence solve further,
$y\left( {{y^2} + 4 - 4y + 4y} \right) - 16 = 0 \\ y{\left( {y - 2} \right)^2} + 4{y^2} - 16 = 0 \\$
Now we will take terms common and simplify,
$y\left( {y - 2} \right)\left( {y - 2} \right) + 4\left( {{y^2} - 4} \right) = 0 \\ \Rightarrow y\left( {y - 2} \right)\left( {y - 2} \right) + 4\left( {y + 2} \right)\left( {y - 2} \right) = 0 \\ \Rightarrow \left( {y - 2} \right)\left( {{y^2} - 2y + 4y + 8} \right) = 0 \\ \Rightarrow \left( {y - 2} \right)\left( {{y^2} + 2y + 8} \right) = 0 \\$
We will equate each of the factors to 0 and find the value of $y$
That is, when $y - 2 = 0$
We get, $y = 2$
Similarly, when we put ${y^2} + 2y + 8 = 0$
We will first calculate $D$ which is $4ac$
Hence, we have
$D = {b^2} - 4ac \\ \Rightarrow D = {\left( 2 \right)^2} - 4\left( 1 \right)\left( 9 \right) \\ \Rightarrow D = 4 - 36 \\ \Rightarrow D = - 34 \\$
Since, $D < 0$, there does not exist any real root corresponding to the factor ${y^2} + 2y + 8$
Therefore, the value of $y$ is 2
Now, we will substitute the value of $y$ in equation (2) to find the value of $x$
Thus,
$x\left( 2 \right) = 4 \\ x = 2 \\$
Hence, the required point is $\left( {2,2} \right)$
Thus, option B is correct.

Note: Parametric equations are a set of equations that represent the variables as a function of an independent variable. Since, there are two sets of equations involving two parameters, two equations will be formed. And the point of the intersection of the equations is the solution of the two equations.