$\text{ A point on the ellipse }4x^2+9y^2=36\text{ . Where the normal is parallel to the line }$$4x-2y-5=0\text{ is: }$

$\text{ Slope of the normal is }2\text{ , so slope of the tangent is }-\frac{1}{2}\text{ . }$

$\text{ Equation of the tangent is }y-y_1=m\left(x-x_1\right)$

$y=-\frac{1}{2}x\pm \sqrt{9\times \frac{1}{4}+4}$

$\Rightarrow x+2y=\pm \mathrm{5......}\left(1\right)$

$\text{ If it touches at }\left(x_1,y_1\right),\text{ then equation of the tangent at }\left(x_1,y_1\right)\text{ to the ellipse is }{S}_1=0$

$\Rightarrow \frac{x{x}_1}{9}+\frac{y{y}_1}{4}=\mathrm{1.......}\left(2\right)$

Comparing (1) and (2) we get

$\frac{x_1}{9}=\frac{y_1}{8}=\pm \frac{1}{5}$

$\Rightarrow (x_1,y_1)=\left(\frac{9}{5},\frac{8}{5}\right)$