The coordinates of the point on the ellipse

(a) (3, 16/3)

According to the question,

$\frac{d y}{d t}=\frac{-d x}{d t}$

$16 x^{2}+9 y^{2}=400$

$\Rightarrow 32 x \frac{d x}{d t}+18 y \frac{d y}{d t}=0$

$\Rightarrow 32 x \frac{d x}{d t}=-18 y \frac{d y}{d t}$

$\Rightarrow 32 x=18 y$

$\Rightarrow x=\frac{9 y}{16}$               .....(1)

Now,

$16\left(\frac{9 y}{16}\right)^{2}+9 y^{2}=400$

$\Rightarrow \frac{81 y^{2}}{16}+9 y^{2}=400$

$\Rightarrow 81 y^{2}+144 y^{2}=6400$

$\Rightarrow 225 y^{2}=6400$

$\Rightarrow y^{2}=\frac{6400}{225}$

$\Rightarrow y=\sqrt{\frac{6400}{225}}$

$\Rightarrow y=\frac{16}{3}$ or $-\frac{16}{3}$

So,

$x=\frac{9}{16} \times \frac{16}{3}$        $\left[\begin{array}{ll}\operatorname{Using} & (1)\end{array}\right]$

Or

$x=-\frac{9}{16} \times \frac{16}{3}$

$\Rightarrow x=3$ or $-3$

So, the required point is $\left(3, \frac{16}{3}\right)$.