Hint: We can compare a given parabola with the general equation of a parabola and obtain the length of the latus rectum as 4. Then we can use the distance formula to get the y coordinate of the ends of the latus rectum. Then we can use the equation of the tangent to the parabola at any point $\left( {{x}_{1}},{{y}_{1}} \right)$, which is given by \[y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)\]. Then, we can substitute the coordinates of the ends of the latus rectum to obtain two equations in x and y. Solving those, we will get the required answer.

Complete step-by-step solution -

The equation of the parabola is given in the question as \[{{y}^{2}}=4x\].

Comparing it with the general equation, \[{{y}^{2}}=4ax\] , we get that \[a=1\].

For the parabola, \[{{y}^{2}}=4ax\], the latus rectum is given by \[x=a\] and the length of the latus rectum is \[4a\]. The focus is $\left( a,0 \right)$.

Therefore, for the parabola \[{{y}^{2}}=4x\], we have the details as,

Latus rectum equation as \[x=1\]

Length of latus rectum $=4$

Focus at \[\left( 1,0 \right)\]

The figure below shows the details,

We have to compute the coordinates of the point $I$. For that, first, we have to find the coordinates of the ends of the latus rectum. The equation \[x=1\] gives the \[x\] coordinate of the latus rectum as $1$. Let the \[y\] coordinate be \[y\].

From the figure, it is clear that the distance \[LS\] is \[2\]. So, using the distance formula,

$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} $

$2=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-y \right)}^{2}}} $

$2=\sqrt{{{y}^{2}}} $

${{y}^{2}}=4 $

$y=\pm 2 $

Since the coordinates of the ends of the latus rectum, \[L\] and \[R\] are given by $\left( 1,y \right)$, we get it as \[\left( 1,2 \right),\left( 1,-2 \right)\] respectively.

The equation of the tangent to parabola at any point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by

\[y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)\ldots \ldots \ldots \left( i \right)\]

Therefore, the equation of the tangent to the end \[L\left( 1,2 \right)\] of the latus rectum can be obtained by substituting \[x=1,y=2,a=1\] in equation \[\left( i \right)\] as,

$2y=2\left( x+1 \right)$

$y=\left( x+1 \right)$

$y-x=1\ldots \ldots \ldots \left( ii \right) $

The equation of the next tangent to the end \[R\left( 1,-2 \right)\] of the latus rectum can be obtained by substituting \[x=1,y=-2,a=1\] in equation \[\left( i \right)\] as,

$-2y=2\left( x+1 \right) $

$-y=\left( x+1 \right) $

$-y-x=1\ldots \ldots \ldots \left( iii \right) $

Adding equations \[\left( ii \right)\] and $\left( iii \right)$, we get

$y-x=1 $

$-y-x=1 $

$-2x=2$

$x=-1$

Substituting \[x=-1\] in equation \[\left( ii \right)\], we get

$y-(-1)=1$

$y+1=1$

$y=0$

Therefore, we get the point of intersection of the tangents at the ends of the latus rectum, $I$ of the parabola \[{{y}^{2}}=4x\] as \[\left( -1,0 \right)\].

The answer obtained is \[\left( -1,0 \right)\].

Note: This question can be solved easily using the fact that the point of intersection of the tangents at the extremities of the latus rectum of a parabola is the foot of the directrix. For the parabola \[{{y}^{2}}=4ax\], the directrix is given by $x=-a$. So, for the parabola \[{{y}^{2}}=4x\] in this question, the directrix is \[x=-1\]. The coordinates of the foot of the directrix are hence \[\left( -1,0 \right)\].