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**Hint:**We will write the value of $y$ from the given equation of family of lines. Use the trigonometric ratios and identities, to simplify the expression. Then, according to the expression obtained, put $x = 2$. If a solution of $y$ exists, then the lines are concurrent and the solution of $\left( {x,y} \right)$ is the point of concurrence.

**Complete step-by-step answer:**We are given that $x{\sec ^2}\theta + y{\tan ^2}\theta - 2 = 0$ represents a family of lines.

We will first find the value of $y$ from the given equation.

$y = \dfrac{{2 - x{{\sec }^2}\theta }}{{{{\tan }^2}\theta }}$

Now, we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{{\text{cos}}\theta }}$

$

y = \dfrac{{2 - x\left( {\dfrac{1}{{{\text{co}}{{\text{s}}^2}\theta }}} \right)}}{{\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}} \\

\Rightarrow y = \dfrac{{2{{\cos }^2}\theta - x}}{{{{\sin }^2}\theta }} \\

\Rightarrow y = 2\dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} - \dfrac{x}{{{{\sin }^2}\theta }} \\

$

Also, \[\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta \] and $\dfrac{1}{{\sin \theta }} = {\text{cosec}}\theta $

$y = 2{\cot ^2}\theta - x{\operatorname{cosec} ^2}\theta $

We want to find if there exists any solution of the above equation

Put $x = 2$ and check if the value of $y$ exists.

$

y = 2{\cot ^2}\theta - 2{\operatorname{cosec} ^2}\theta \\

\Rightarrow y = 2\left( {{{\cot }^2}\theta - {{\operatorname{cosec} }^2}\theta } \right) \\

$

And from the identity, we have ${\operatorname{cosec} ^2}\theta - {\cot ^2}\theta = 1$

Then,

$

y = 2\left( { - 1} \right) \\

\Rightarrow y = - 2 \\

$

Hence, the family of lines is concurrent at $\left( { - 2,2} \right)$

**Thus, option C is correct.**

**Note:**A set of lines is called a family of lines having some common parameter. Three or more lines are called concurrent if they pass through a common point. The point from which the lines pass is known as point of occurrence.

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