Question

A curve is represented by the equation, $ x = {\sec ^2}t $ and $ y = \cot t $ , where $ t $ is a parameter. If the tangent at the point on the curve where $ t = \dfrac{\pi }{4} $ meets the curve again at the point Q then $ |PQ| $ is Equal to ?
A. $ \dfrac{{5\sqrt 3 }}{2} $
B. $ \dfrac{{5\sqrt 5 }}{2} $
C. $ \dfrac{{2\sqrt 5 }}{3} $
D. $ \dfrac{{3\sqrt 5 }}{2} $

Answer 1

Hint: In these types of questions it is expected from the student to know the basic equations of the curves in different forms.Such questions are tricky and therefore students must be able to identify the terms associated with the variables in the questions.
Students must be able to comprehend these variables and establish a relationship between them to solve the question.
Should know the use of trigonometric expressions involved.

Complete step-by-step answer:
let us understand the given data,
 $ x = {\sec ^2}t $
And
 $ y = \cot t $
Represent a curve.
Let us differentiate them with respect to $ t $
Thus, we get
 $ \dfrac{{dx}}{{dt}} = 2\sec t(\sec t.\tan t) $
And
 $ \dfrac{{dy}}{{dt}} = - \cos e{c^2}t $
Now, we know that derivative of y with respect to x is given by $ \dfrac{{dy}}{{dx}} $
Thus,
Modifying it to get to the answer.
 $ \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} $
Substituting the values of $ \dfrac{{dy}}{{dt}}and\dfrac{{dx}}{{dt}} $
 $
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \cos e{c^2}t}}{{2\sec t(\sec t.\tan t)}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{{\cos }^2}t.\cos t}}{{2{{\sin }^2}t.\sin t}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{{\cos }^3}t}}{{2{{\sin }^3}t}} \\
  $
In the question we have been also provided with $ t $ which is a parameter.
Also, the tangent at the point P on the curve where $ t = \dfrac{\pi }{4} $ meets the curve again at the point Q.
Thus,
 $ \mathop {\dfrac{{dy}}{{dx}}}\nolimits_{t = \dfrac{\pi }{4}} = - \dfrac{{{{\cos }^3}(\dfrac{\pi }{4})}}{{{{\sin }^3}(\dfrac{\pi }{4})}} $
 $
  \mathop { \Rightarrow \dfrac{{dy}}{{dx}}}\nolimits_{t = \dfrac{\pi }{4}} = - \dfrac{{{{(\dfrac{1}{{\sqrt 2 }})}^3}}}{{2 \times {{(\dfrac{1}{{\sqrt 2 }})}^3}}} \\
  \mathop { \Rightarrow \dfrac{{dy}}{{dx}}}\nolimits_{t = \dfrac{\pi }{4}} = - \dfrac{1}{2} \\
  $
The above equation gives the slope of the tangent.
Now, as mentioned in the question this tangent meets the curve again. Say at point $ {t_1} $
With same conditions as $ t $ at $ \dfrac{\pi }{4} $
Then we have,
 $ $ $ x = {\sec ^2}t $
And
 $ y = \cot t $
At $ \dfrac{\pi }{4} $ the above results give us,
 $
  x = {\sec ^2}t \\
   \Rightarrow x = {\sec ^2}\dfrac{\pi }{4} \\
   \Rightarrow x = 2 \\
  $
And
 $
  y = \cot t \\
   \Rightarrow y = \cot \dfrac{\pi }{4} \\
   \Rightarrow y = 1 \\
  $
Let us consider the equation of the tangent,
 $ (y - 1) = - \dfrac{1}{2}(x - 2) $
 $ \Rightarrow x + 2y = 4 $
Next step is to solve the two tangents,
 $ \Rightarrow x + 2y = 4 $ and $ \Rightarrow {y^2} = \dfrac{1}{{x - 1}} $
We get the point Q as $ (5, - \dfrac{1}{2}) $
Now, we have two point \[ - (5, - \dfrac{1}{2})\] and $ (2,1) $
Thus we can solve to find $ |PQ| $ by distance formula.
 $
  |PQ| = \sqrt {{{(5 - 2)}^2} + {{( - \dfrac{1}{2} - 1)}^2}} \\
   \Rightarrow |PQ| = \sqrt {{{(3)}^2} + {{( - \dfrac{3}{2})}^2}} \\
   \Rightarrow |PQ| = \sqrt {9 + \dfrac{9}{2}} \\
   \Rightarrow |PQ| = \sqrt {\dfrac{{18 + 9}}{2}} \\
   \Rightarrow |PQ| = \sqrt {\dfrac{{27}}{2}} \\
   \Rightarrow |PQ| = \sqrt {\dfrac{{9 \times 3}}{2}} \\
   \Rightarrow |PQ| = 3\sqrt {\dfrac{3}{2}} \\
   \Rightarrow |PQ| = 3\sqrt {\dfrac{{3 + 2}}{{2 + 2}}} \\
   \Rightarrow |PQ| = 3\sqrt {\dfrac{5}{4}} \\
   \Rightarrow |PQ| = \dfrac{{3\sqrt 5 }}{2} \\
  $
So, the correct answer is “Option D”.

Note: In these types of examples the chances are that students can make mistakes in calculations so it is advised that they go step by step and not hurry considering the signs of the equations.
It is important to understand what the equation demands and based on the practical knowledge we derive the equations of tangents based on the curve.