The ellipse is
$$\frac{x^2}{15^2}+\frac{y^2}{9^2}=1$$
Hence
$$a=15,\quad b=9$$
The focal distance is
$$c=\sqrt{a^2-b^2}$$
$$=\sqrt{225-81}$$
$$=\sqrt{144}$$
$$=12$$
Therefore the foci are
$$S(12,0),\quad S'(-12,0)$$
Let $$P=(x,y)$$ be any point on the ellipse.
Since $$P$$ lies on the ellipse,
$$PS+PS'=2a=30$$
Now consider triangle
$$\triangle PSS'$$
Its sides are
$$PS,\quad PS',\quad SS'=24$$
The incenter divides the angle bisectors in the ratio of the adjacent sides.
Using barycentric coordinates of the incenter,
$$I=\frac{24P+PS'\,S+PS\,S'}{24+PS+PS'}$$
Since
$$PS+PS'=30$$
the denominator becomes
$$24+30=54$$
Hence the y-coordinate of the incenter is
$$y_I=\frac{24y}{54}$$
$$=\frac49\,y$$
Therefore maximizing $$y_I$$ is equivalent to maximizing $$y$$ on the ellipse.
From $$\frac{x^2}{225}+\frac{y^2}{81}=1$$ the maximum value of $$y$$ is $$9$$ attained at $$(0,9)$$
Therefore $$y_I^{\max}=\frac49\cdot 9$$
$$=4$$
Final Answer :
$$4$$
91- 6303239042
