Ans : (B, D)
Let \(\left(\frac{1}{2} \cos \theta, \frac{1}{3} \sin \theta\right)\) be a point on \(4 x^{2}+9 y^{2}=1\), so equation of \(\operatorname{tangent}\left(\frac{1}{2} \cos \theta, \frac{1}{3} \sin \theta\right)\) is
\(2 x \cos \theta+3 y \sin \theta=1\)
equating slope with \(8 x=9 y\)
\(\frac{-2 \cos \theta}{3 \sin \theta}=\frac{8}{9} \Rightarrow \tan \theta=-\frac{3}{4}\)
Hence either \(\cos \theta=-\frac{4}{5}, \sin \theta=\frac{3}{5}\)
or \(\cos \theta=\frac{4}{5}, \sin \theta=-\frac{3}{5}\)
so the points are \(\left(-\frac{2}{5}, \frac{1}{5}\right)\) or \(\left(\frac{2}{5},-\frac{1}{5}\right)\)
