A point on the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at a distance equal to the mean of lengths of the semi-major and semi-minor axis from the centre, is

Let $P(4\cos \theta ,3\sin \theta )$ be a point on the given ellipse such that its distance from the centre $(0, 0)$ of the ellipse is equal to the mean of the lengths of the semi-major and semi-minor axes i.e.

$\begin{array}{l}OP=\frac{4+3}{2}\\ \Rightarrow \sqrt{16{\cos }^2\theta +9{\sin }^2\theta }=\frac{7}{2}\\ \Rightarrow 7{\cos }^2\theta +9=\frac{49}{4}\\ \Rightarrow {\cos }^2\theta =\frac{13}{28}\end{array}$

$\Rightarrow \cos \theta =\pm \sqrt{\frac{13}{28}}$ and $\sin \theta =\pm \sqrt{\frac{15}{28}}$

$\Rightarrow \cos \theta =\pm \frac{\sqrt{91}}{14}$ and $\sin \theta =\pm \frac{\sqrt{105}}{14}$

Hence, the required points are given by

$P\left(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{14}\right)$