Question

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 by
A) (4,5)
B) (2,2)
C) (-2,4)
D) (1,2)

Answer 1

We will first eliminate the t form $x = {t^2} + 1$ , $y = 2t$ by multiplying equation 1 by 4 and then we will square given y. Now we will substitute $4{t^2}$into ${y^2}$ we will get our equation of elimination of t, then we will put values $x = 2s$ and $y = \dfrac{2}{s}$ into the equation, from which we will find the value of s and on further simplification we will reach to our answer.

Complete step by step solution:
We will start with eliminating $t$from
$x = {t^2} + 1$ …($1$)
$y = 2t$ …(2)
We will be going to multiply equation (1) by 4 and going to square equation (2)
We get
$\Rightarrow 4x = 4{t^2} + 4$ …(3)
$\Rightarrow {y^2} = 4{t^2}$
Substituting the value of $4{t^2}$ in equation (3), we get,
$\Rightarrow 4x = {y^2} + 4$
On rearranging we get,
$\Rightarrow {y^2} = 4x - 4$
On Substituting $x = 2s$ and $y = \dfrac{2}{s}$ in ${y^2} = 4x - 4$ , we get,
$\Rightarrow {\left( {\dfrac{2}{s}} \right)^2} = 4 \times 2s - 4$
On simplification we get,
$\Rightarrow \left( {\dfrac{4}{{{s^2}}}} \right) = 8s - 4$
On cross multiplication we get,
$\Rightarrow 4 = {s^2}(8s - 4)$
On simplification we get,
$\Rightarrow 1 = 2{s^3} - {s^2}$
On subtracting 1 from both sides we get,
$\Rightarrow 2{s^3} - {s^2} - 1 = 0$
We can split further into
$\Rightarrow (s - 1)(2{s^2} + s + 1) = 0$
So, we have,
$\Rightarrow (s - 1) = 0$ and $(2{s^2} + s + 1) = 0$
As $(2{s^2} + s + 1) = 0$ is not possible
So, we have
$\Rightarrow (s - 1) = 0$
Hence, we have,
$\Rightarrow s = 1$
On substituting $s = 1$ in
$x = 2s$
$y = \dfrac{2}{s}$
After substituting the value, we will get
$x = 2 \times 1$
$y = \dfrac{2}{1}$
These are the values we get for the x and y
$x = 2$
$y = 2$
Hence the answer is (2,2).

So, the answer is B.

Note:
For intersection of both the curve we must have
${t^2} + 1 = 2s$ and $2t = {s^2} \Rightarrow ts = 1$
We will apply the squaring method to find the point of the intersection
$\Rightarrow \left( {t - 1} \right)({t^2} + t + 2) = 0$
By calculating further, we will get that
$\Rightarrow t = 1$
is the only root
Therefore, the point of intersection is (2,2).