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
$p^2=x^2+y^2$
$$\begin{gathered} =a^2{\sin }^{2}t+b^2{\sin }^2\left(\frac{\text{ at }}{b}\right)-2ab\sin t\cdot \sin \left(\frac{\text{ at }}{b}\right)+a^2{\cos }^{2}t+b^2{\cos }^2\left(\frac{\text{ at }}{b}\right)-2ab\cos t\cdot \cos \left(\frac{\text{ at }}{b}\right) \\ <br/>=a^2+b^2-2ab\left[\sin t\cdot \sin \left(\frac{\text{ at }}{b}\right)+\cos t\cdot \cos \left(\frac{\text{ at }}{b}\right)\right] \\ <br/>=a^2+b^2-2ab\left[\cos \left(t-\frac{at}{b}\right)\right] \end{gathered}$$
Now
$\frac{d{p}^2}{dt}$
$<br/>=-2ab\sin \left(t-\frac{at}{b}\right)\left(1-\frac{a}{b}\right)<br/>$
$<br/>=0<br/>$
Or
$<br/>\sin \left(t-\frac{\text{ at }}{b}\right)=0<br/>$
Hence
$<br/>\cos \left(t-\frac{at}{b}\right)=1<br/>$
Substituting in the expression of $p^2$ we get
$p^2=a^2+b^2-2ab$
$=(a-b{)}^2$
Hence
$p_{\max }=\sqrt{(a-b{)}^2}$
$=a-b$