Answer

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**Hint:**Here, we have to find the equation of parabola. First, we will find the length of the latus rectum and then equate it to the formula of the latus rectum and find the coordinate of the focus. Using this we will find the coordinate of a vertex. Then substitute the coordinate of a vertex in the general equation of parabola and simplify it further to get the required answer.

**Formula Used:**

We will use the following formulas:

1. Distance between two points by using the given formula \[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \] where \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are the coordinates of two points.

2. Length of the Latus Rectum is given by the formula Length of the Latus Rectum\[ = 4a\] where \[a\] is the coordinate of focus.

3. The standard equation of a parabola is given by \[{x^2} = 4ay\] where \[S(0,a)\] is the coordinate of the focus.

**Complete Step by Step Solution:**

We are given the endpoints of the latus rectum of a parabola are \[\left( { - 3,1} \right)\] and \[\left( {1,1} \right)\].

By using the distance formula \[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \] , we will find the length of the latus rectum with the coordinates of the ends of the latus rectum. Therefore, we get

Length of the latus rectum \[ = \sqrt {{{\left( { - 3 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} \]

Adding and subtracting the terms inside the bracket, we get

\[ \Rightarrow \] Length of the latus rectum \[ = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 0 \right)}^2}} \]

Squaring and adding the terms, we get

\[ \Rightarrow \] Length of the latus rectum \[ = \sqrt {16} \]

Now taking the square root, we get

\[ \Rightarrow \] Length of the latus rectum \[ = \pm 4\]

We know that the length of the latus rectum cannot be negative, so the length is 4

Now, we know that the length of the Latus Rectum is given by the formula \[ = 4a\].

So, by equating the length of the latus rectum to \[4a\], we get

\[4a = 4\]

Dividing by 4 on both the sides, we get

\[ \Rightarrow a = 1\]

The endpoints of the latus rectum are of the form \[\left( { \pm 2a,a} \right)\].

So, the equation of the parabola is given by \[{x^2} = 4ay\]

Since the given is a parabola, we have to find the coordinate of a vertex.

\[\begin{array}{l}2a = - 3 + 1\\ \Rightarrow 2a = - 2\end{array}\]

Dividing both side by 2, we get

\[ \Rightarrow a = - 1\]

\[\begin{array}{l}a = 1 - 1\\ \Rightarrow a = 0\end{array}\]

So, the coordinate of the vertex \[(h,k)\]is \[( - 1,0)\]

When the parabola is not at origin, then the equation of the parabola is \[{\left( {x - h} \right)^2} = 4a\left( {y - k} \right)\]

Substituting the known values, we get

\[ \Rightarrow {\left( {x - ( - 1)} \right)^2} = 4(1)(y - 0)\]

By simplifying the terms, we get

\[ \Rightarrow {\left( {x + 1} \right)^2} = 4y\]

Therefore, the equation of the parabola is \[{\left( {x + 1} \right)^2} = 4y\].

Now, we will draw a parabola using the above equation. This parabola will open upwards.

**Hence option A is the correct answer.**

**Note:**

We know that the latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and its endpoints lie on the parabola. A parabola is a curve where any point is at an equal distance from a fixed point (the focus ) and a fixed straight line (the directrix ).

A parabola is symmetric with its axis. If the equation has a \[{y^2}\] term, then the axis of symmetry is along the \[x\]-axis and if the equation has an \[{x^2}\] term, then the axis of symmetry is along the \[y\]-axis. So, the given equation of Parabola is open upwards.

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