Question

The equations $x=a\cos \theta +b\sin \theta \text{ and }y=a\sin \theta -b\cos \theta $ represent
(a) Circle
(b) A parabola
(c) A line
(d) An ellipse

Answer 1

Hint: Squaring both terms and then adding them then apply a trigonometric formula.

Complete step-by-step answer:
 Given equations are:
(1)$x=a\cos \theta +b\sin \theta $
Squaring both sides; it will give
${{x}^{2}}={{(a\cos \theta +b\sin \theta )}^{2}}$
\[{{x}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\cos \theta .\sin \theta \]
(2) $y=a\sin \theta -b\cos \theta $
Squaring both sides; it will give
${{y}^{2}}={{(a\sin \theta -b\cos \theta )}^{2}}$
${{y}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta .\sin \theta $
Now adding $\left( {{x}^{2}}+{{y}^{2}} \right)$, we get
\[\begin{align}
  & \text{ }{{x}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{2ab\cos \theta .\sin \theta } \\
 & \underline{(+)\text{ }{{y}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -{2ab\sin \theta .\cos \theta }} \\
\end{align}\]
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta $
Rearranging the term, we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta $
Taking out the common term, we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+{{b}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )$
Now we know, $\left[ {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right]$, so the above expression can be written as,
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}\times 1+{{b}^{2}}\times 1$
$\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$
This represents an equation of circle with origin as centre and $(\sqrt{{{a}^{2}}+{{b}^{2}}})$ as the radius.
Hence the correct answer is option (a).
Answer is option (a)

Note: Whenever this type of question is given, that involves sine and cosine functions, squaring is must and then add the result. This is the easiest way to solve it.
Suppose that we consider the operation $\dfrac{x}{y}$ , we get a different result.