The maximum distance from origin of a point on the curve $x=a \sin t-b \sin \left(\frac{a t}{b}\right)$ $y=a…
The maximum distance from origin of a point on the curve $x=a \sin t-b \sin \left(\frac{a t}{b}\right)$
$y=a \cos t-b \cos \left(\frac{a t}{b}\right)$, both $a, b>0$ is
a - b
a + b
$\sqrt{a^2+b^2}$
$\sqrt{a^2-b^2}$
Solution
Distance of origin from $(x, y)=\sqrt{x^2+y^2}$
$
=\sqrt{a^2+b^2-2 a b \cos \left(t-\frac{a t}{b}\right)}=\sqrt{a^2+b^2-2 a b}\left[\because \max \cdot \cos \left(t-\frac{a t}{b}\right)=1\right]=a-b
$