Relation Between Semi-Axes and Focal Distance

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 :

$$4$$