question_answer

If \[x=3\text{ }sin\text{ }\theta \] and \[y=4\text{ }cos\,\theta \], find the value of \[\sqrt{16{{x}^{2}}+9{{y}^{2}}}\].

Answer:

\[x=3\text{ }sin\text{ }\theta \]

\[\Rightarrow \]            \[{{x}^{2}}=9\,\,{{\sin }^{2}}\theta \]

\[\,{{\sin }^{2}}\theta =\frac{{{x}^{2}}}{9}\]                                                         ?(i)

And                  \[y=4\,\,\cos \,\,\theta \]

\[{{y}^{2}}=16\,\,{{\cos }^{2}}\,\,\theta \]

\[{{\cos }^{2}}\theta =\frac{{{y}^{2}}}{16}\]                                                ?(ii)

On adding eq. (i) and eq. (ii)

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}\]

\[1=\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}\]

\[1=\frac{16{{x}^{2}}+9{{y}^{2}}}{144}\]

\[16{{x}^{2}}+9{{y}^{2}}=144\]

\[\sqrt{16{{x}^{2}}+9{{y}^{2}}}=\sqrt{144}\]

\[\sqrt{16{{x}^{2}}+9{{y}^{2}}}=12\]