Question

The maximum distance from $(0,0)$ to a point on the curve $x=a\sin t-b\sin \left(\frac{at}{b}\right)$ and $y=a\cos t-b\cos \left(\frac{at}{b}\right)$ for 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

The correct option is A $a-b$

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