Key Concepts and Formulas

• Distance Formula: The distance $D$ between a point $(x, y)$ and the origin $(0, 0)$ is given by $D = \sqrt{x^2 + y^2}$.
• Trigonometric Identity: $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
• Maximization: To find the maximum of $D = \sqrt{x^2 + y^2}$, it is equivalent to find the maximum of $D^2 = x^2 + y^2$.

Step-by-Step Solution

Step 1: Express the distance squared in terms of t

We want to find the maximum value of the square of the distance from the origin, $D^2 = x^2 + y^2$. We are given $x = a\sin t - b\sin \left( {{{at} \over b}} \right)$ and $y = a\cos t - b\cos \left( {{{at} \over b}} \right)$. Therefore, $D^2 = \left(a\sin t - b\sin \left( {{{at} \over b}} \right)\right)^2 + \left(a\cos t - b\cos \left( {{{at} \over b}} \right)\right)^2$

Step 2: Expand the squares

Expanding the squares, we get: $D^2 = a^2\sin^2 t - 2ab\sin t \sin \left( {{{at} \over b}} \right) + b^2\sin^2 \left( {{{at} \over b}} \right) + a^2\cos^2 t - 2ab\cos t \cos \left( {{{at} \over b}} \right) + b^2\cos^2 \left( {{{at} \over b}} \right)$

Step 3: Simplify using trigonometric identities

We can group terms and use the identity $\sin^2 \theta + \cos^2 \theta = 1$: $D^2 = a^2(\sin^2 t + \cos^2 t) + b^2\left(\sin^2 \left( {{{at} \over b}} \right) + \cos^2 \left( {{{at} \over b}} \right)\right) - 2ab\left(\sin t \sin \left( {{{at} \over b}} \right) + \cos t \cos \left( {{{at} \over b}} \right)\right)$ $D^2 = a^2(1) + b^2(1) - 2ab\left(\cos t \cos \left( {{{at} \over b}} \right) + \sin t \sin \left( {{{at} \over b}} \right)\right)$

Step 4: Apply the cosine subtraction formula

Using the cosine subtraction formula, $\cos(A - B) = \cos A \cos B + \sin A \sin B$, with $A = t$ and $B = \frac{at}{b}$, we have: $D^2 = a^2 + b^2 - 2ab\cos\left(t - \frac{at}{b}\right)$ $D^2 = a^2 + b^2 - 2ab\cos\left(t\left(1 - \frac{a}{b}\right)\right)$ $D^2 = a^2 + b^2 - 2ab\cos\left(\frac{b-a}{b}t\right)$

Step 5: Maximize the distance squared

To maximize $D^2$, we need to minimize the term $\cos\left(\frac{b-a}{b}t\right)$. The minimum value of the cosine function is -1. Therefore, the maximum value of $D^2$ is: $D^2_{max} = a^2 + b^2 - 2ab(-1) = a^2 + b^2 + 2ab = (a + b)^2$

Step 6: Find the maximum distance

Taking the square root to find the maximum distance $D_{max}$, we get: $D_{max} = \sqrt{(a + b)^2} = a + b$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding and simplifying expressions.
• Trigonometric Identities: Memorize and recognize trigonometric identities to simplify expressions effectively.
• Maximization/Minimization: Remember that maximizing $D$ is equivalent to maximizing $D^2$ when $D$ is non-negative. Also remember the range of cosine function [-1,1].

Summary

We found the maximum distance from the origin of a point on the given curve by first expressing the squared distance as a function of the parameter t. Then, using trigonometric identities, we simplified the expression and found the value of t that maximizes the squared distance. Finally, we took the square root to find the maximum distance, which is $a+b$.

Final Answer The final answer is \boxed{a+b}, which corresponds to option (B).