Question

“The segment of the normal to the curve $x = 2a\sin t + a\sin t{\cos ^2}t;y = - a{\cos ^3}t$ contained between the coordinate axes is equal to 2a”. the statement is:
A.True
B.False

Answer 1

Hint: Here, we are required to first find the slope of curve and then we will apply the chain rule. Then we will find the slope of the normal to the curve. Then, we will find the equation of the normal using the slope intercept form of the straight lines. Then for the intercepts, we will put y= 0 and x= 0 for x-intercept and y-intercept respectively. We will find the segment by using the distance formula. If it appears to be 2a then we will say that the given statement is true, otherwise, false.

Complete step-by-step answer:
First, we are given the equation of curve $x = 2a\sin t + a\sin t{\cos ^2}t;y = - a{\cos ^3}t$
For finding the slope of the given curve, we have
$
  \dfrac{{dx}}{{dt}} = 2a\cos t + a{\cos ^3}t - 2a{\sin ^2}t\cos t \\
   \Rightarrow \dfrac{{dx}}{{dt}} = a\cos t\left( {2 + {{\cos }^2}t - 2{{\sin }^2}t} \right) \\
 $
Also,
$
  \dfrac{{dy}}{{dt}} = - 3a{\cos ^2}t\left( { - \sin t} \right) \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 3a{\cos ^2}t\sin t \\
 $
We need to find $\dfrac{{dy}}{{dx}}$. Using chain rule, we can write it as:
$
  \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \cdot \dfrac{{dt}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{3a{{\cos }^2}t\sin t}}{{a\cos t\left( {2 + {{\cos }^2}t - 2{{\sin }^2}t} \right)}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{3\cos t\sin t}}{{2 + {{\cos }^2}t + {{\sin }^2}t - 3{{\sin }^2}t}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{3\cos t\sin t}}{{3\left( {1 - {{\sin }^2}t} \right)}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos t\sin t}}{{{{\cos }^2}t}} = \tan t \\
 $
Now, we need to find the slope of the normal= $ - \dfrac{{dx}}{{dy}} = - \dfrac{1}{{\tan t}} = - \cot t$
Therefore, the equation of normal using the slope intercept form of the equation of straight lines, we get
$y - {y_1} = m\left( {x - {x_1}} \right)$
$ \Rightarrow y + a{\cos ^3}t = - \cot t\left( {x - 2a\sin t - a\sin t{{\cos }^2}t} \right)$
Therefore, now we are only left with the intercepts i.e., x- intercept and the y- intercept.
For x- intercept, we will put y=0 and for y-intercept, we will put x= 0.
For y= 0, the above equation becomes:
$0 + a{\cos ^3}t = - \cot t\left( {x - 2a\sin t - a\sin t{{\cos }^2}t} \right)$
 $
   \Rightarrow - \dfrac{{a{{\cos }^3}t}}{{\cos t}}\sin t = x - 2a\sin t - a\sin t{\cos ^2}t \\
   \Rightarrow x = 2a\sin t + a{\cos ^2}t\sin t - a\sin t{\cos ^2}t \\
   \Rightarrow x = 2a\sin t \\
 $
Similarly, for x= 0, the equation of normal becomes:
$
  y + a{\cos ^3}t = 2a\sin t\cot t + a\sin t{\cos ^2}t\cot t \\
   \Rightarrow y + a{\cos ^3}t = 2a\cos t + a{\cos ^3}t \\
   \Rightarrow y = 2a\cos t \\
 $
Therefore, we get the x-coordinate (2asint, 0) and y-coordinate (0, 2acost).
Now, when we have the values of x- intercept and y- intercept, so the distance between them can be calculated by using the distance formula i.e.,
Distance between x- intercept and y- intercept = $\sqrt {{{\left( {2a\sin t - 0} \right)}^2} + {{\left( {0 - 2a\cos t} \right)}^2}} $
$ \Rightarrow $distance = $\sqrt {4{a^2}{{\sin }^2}t + 4{a^2}{{\cos }^2}t} = 2a\sqrt {{{\sin }^2}t + {{\cos }^2}t} $
$ \Rightarrow $distance = 2a.
Therefore, the given statement is true and option(A) is correct.

Note: In such questions, we suppose the given statement to be merely a question and we derive the solution step by step (not claiming that we need to prove something). In the end if the answer matches the solution obtained, then only we verify the answer.