Answer
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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.
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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