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Question: Family of the lines \(x{\sec ^2}\theta + y{\tan ^2}\theta - 2 = 0\), for different real \(\theta \),...

Family of the lines xsec⁡2θ+ytan⁡2θ−2=0x{\sec ^2}\theta + y{\tan ^2}\theta - 2 = 0, for different real θ\theta , is:
(A) Not concurrent
(B) Concurrent at (1,1)\left( {1,1} \right)
(C) Concurrent at (2,−2)\left( {2, - 2} \right)
(D) Concurrent at (−2,2)\left( { - 2,2} \right)

Explanation

Solution

We will write the value of yy 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=2x = 2. If a solution of yy exists, then the lines are concurrent and the solution of (x,y)\left( {x,y} \right) is the point of concurrence.

Complete step-by-step answer:
We are given that xsec⁡2θ+ytan⁡2θ−2=0x{\sec ^2}\theta + y{\tan ^2}\theta - 2 = 0 represents a family of lines.
We will first find the value of yy from the given equation.
y=2−xsec⁡2θtan⁡2θy = \dfrac{{2 - x{{\sec }^2}\theta }}{{{{\tan }^2}\theta }}
Now, we know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and sec⁡θ=1cosθ\sec \theta = \dfrac{1}{{{\text{cos}}\theta }}
y=2−x(1cos2θ)sin⁡2θcos⁡2θ ⇒y=2cos⁡2θ−xsin⁡2θ ⇒y=2cos⁡2θsin⁡2θ−xsin⁡2θ  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, cos⁡θsin⁡θ=cot⁡θ\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta and 1sin⁡θ=cosecθ\dfrac{1}{{\sin \theta }} = {\text{cosec}}\theta
y=2cot⁡2θ−xcosec⁡2θy = 2{\cot ^2}\theta - x{\operatorname{cosec} ^2}\theta
We want to find if there exists any solution of the above equation
Put x=2x = 2 and check if the value of yy exists.
y=2cot⁡2θ−2cosec⁡2θ ⇒y=2(cot⁡2θ−cosec⁡2θ)  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 cosec⁡2θ−cot⁡2θ=1{\operatorname{cosec} ^2}\theta - {\cot ^2}\theta = 1
Then,
y=2(−1) ⇒y=−2  y = 2\left( { - 1} \right) \\\ \Rightarrow y = - 2 \\\
Hence, the family of lines is concurrent at (−2,2)\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.