Question

Locus of the point of the intersection of the lines \[xcos\theta = y\] and \[cot\theta = a\] is
A) ${x^2} + {y^2} = 2{a^2}$
B) ${x^2} + {y^2} - ax = 0$
C) ${y^2} = 4ax$
D) ${x^2} = {a^2} + {y^2}$

Answer 1

Hint: In these questions, there are two lines given in the question. Firstly we will equate these two lines to get the value of y and denote it as equation (1) then take another equation from the question and with the help of that we can easily get our answer.

Complete step-by-step answer:
There are two lines $x\cos \theta $ and $y\cot \theta $
Now, equating these two lines as
$
  \therefore x\cos \theta = y\cot \theta \\
   \Rightarrow x\cos \theta = y\dfrac{{\cos \theta }}{{\sin \theta }} \\
  \therefore y = x\sin \theta {\text{ for above equation to be held}} \\
$
Also, $xCos\theta = a$
Let $
 {\text{Equation1 is y = xSin}}\theta \\
  {\text{and Equation 2 is xCos}}\theta {\text{ = a}} \\
$
$\therefore {\text{Squaring and adding equation 1 and equation 2 , we get}}$
$ \Rightarrow {y^2} + {a^2} = {x^2}Si{n^2}\theta + {x^2}Co{s^2}\theta $
$ \Rightarrow {x^2}(Si{n^2}\theta + Co{s^2}\theta ) = {a^2} + {y^2}$
$ \Rightarrow {x^2}(1) = {a^2} + {y^2}$ [$\because Si{n^2}\theta + Co{s^2}\theta = 1$ ]
$ \Rightarrow {x^2} = {a^2} + {y^2}$
Which is the same as Option D.
$\therefore {\text{ Option D is the correct answer}}$

 
Note: Another way for solving these types of problems is by first assembling the point of intersection of given lines, then making equations suitable to the point performing simple calculations, we get our required answer.