Question

The maximum distance from the origin of a point on the curve $x = a$ sin $t- b$ sin$\left(\frac{at}{b}\right),y=a$ cos $t-b$ cos $\left(\frac{at}{b}\right).$ both $a, b >0, is$

• A. $a-b$
• B. $a+b$
• C. $\sqrt{a^{2}+b^{2}}$
• D. $\sqrt{a^{2}-b^{2}}$

Answer 1

Answer: (A)

Let the distance from origin be
$= p $
$=\sqrt{ x ^2+ y ^2}$
Now
$p ^2= x ^2+ y ^2$
$=a^2 \sin ^2 t+b^2 \sin ^2\left(\frac{\text { at }}{b}\right)-2 a b \sin t \cdot \sin \left(\frac{\text { at }}{b}\right)+a^2 \cos ^2 t+b^2 \cos ^2\left(\frac{\text { at }}{b}\right)-2 a b \cos t \cdot \cos \left(\frac{\text { at }}{b}\right) \\
=a^2+b^2-2 a b\left[\sin t \cdot \sin \left(\frac{\text { at }}{b}\right)+\cos t \cdot \cos \left(\frac{\text { at }}{b}\right)\right] \\
=a^2+b^2-2 a b\left[\cos \left(t-\frac{a t}{b}\right)\right]$
Now
$\frac{ dp ^2}{ dt }$
$
=-2 a b \sin \left(t-\frac{a t}{b}\right)\left(1-\frac{a}{b}\right)
$
$
=0
$
Or
$
\sin \left(t-\frac{\text { at }}{b}\right)=0
$
Hence
$
\cos \left( t -\frac{ at }{ b }\right)=1
$
Substituting in the expression of $p ^2$ we get
$p ^2= a ^2+ b ^2-2 ab $
$=( a - b )^2$
Hence
$p _{\max }=\sqrt{( a - b )^2} $
$= a - b$