Foci of a Horizontal Ellipse

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 :

$$4$$

1. Find the Foci and Eccentricity of Hyperbola $$H$$

The given hyperbola is $$H:\frac{x^2}{4}-\frac{y^2}{1}=1.$$

Here, $$a_H^2=4$$ and $$b_H^2=1.$$

Its eccentricity is $$e_H=\sqrt{1+\frac{b_H^2}{a_H^2}}=\sqrt{1+\frac14}=\frac{\sqrt5}{2}.$$

The positive focus of the hyperbola is $$(a_He_H,0).$$

Hence, $$\text{Focus}=\left(2\cdot\frac{\sqrt5}{2},0\right)=(\sqrt5,0).$$

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2. Find the Equation of the Ellipse $$E$$

Since the ellipse $$E$$ is confocal with $$H$$, both have the same foci.

Therefore, $$a_Ee_E=\sqrt5.$$

We are given that the eccentricity of the ellipse is the reciprocal of the eccentricity of the hyperbola:

$$e_E=\frac1{e_H}=\frac{2}{\sqrt5}.$$

Substituting into $$a_Ee_E=\sqrt5,$$

$$a_E\left(\frac2{\sqrt5}\right)=\sqrt5.$$

Hence,

$$a_E=\frac52,\qquad a_E^2=\frac{25}{4}.$$

Now,

$$b_E^2=a_E^2(1-e_E^2)=\frac{25}{4}\left(1-\frac45\right)=\frac{25}{4}\cdot\frac15=\frac54.$$

Thus the ellipse is $$\frac{x^2}{\frac{25}{4}}+\frac{y^2}{\frac54}=1.$$

Equivalently,

$$\frac{4x^2}{25}+\frac{4y^2}{5}=1.$$

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3. Find the Point of Intersection $$P$$

From the hyperbola,

$$y^2=\frac{x^2}{4}-1.$$

Substitute into the ellipse equation:

$$\frac{4x^2}{25}+\frac45\left(\frac{x^2}{4}-1\right)=1.$$

Simplifying,

$$\frac{4x^2}{25}+\frac{x^2}{5}-\frac45=1.$$

Therefore,

$$\frac{4x^2+5x^2}{25}=1+\frac45=\frac95.$$

Hence,

$$\frac{9x^2}{25}=\frac95.$$

Thus,

$$x^2=5.$$

Since $$P$$ lies in the first quadrant,

$$x=\sqrt5.$$

Now,

$$y^2=\frac54-1=\frac14.$$

Therefore,

$$y=\frac12.$$

Hence,

$$P=\left(\sqrt5,\frac12\right).$$

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4. Find the Tangent and Directrix to Locate $$Q$$

The directrix corresponding to the positive focus of the ellipse is $$x=\frac{a_E}{e_E}.$$

Substituting the values,

$$x=\frac{\frac52}{\frac2{\sqrt5}}=\frac{5\sqrt5}{4}.$$

Therefore, the directrix is $$x=\frac{5\sqrt5}{4}.$$

The tangent to $$\frac{4x^2}{25}+\frac{4y^2}{5}=1$$ at $$\left(\sqrt5,\frac12\right)$$ is

$$\frac{4x(\sqrt5)}{25}+\frac{4y\left(\frac12\right)}5=1.$$

That is,

$$\frac{4\sqrt5\,x}{25}+\frac{2y}{5}=1.$$

To find $$Q$$, substitute $$x=\frac{5\sqrt5}{4}$$ into the tangent equation:

$$\frac{4\sqrt5}{25}\left(\frac{5\sqrt5}{4}\right)+\frac{2y}{5}=1.$$

This gives

$$1+\frac{2y}{5}=1.$$

Hence,

$$y=0.$$

Therefore,

$$Q=\left(\frac{5\sqrt5}{4},0\right).$$

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5. Calculate the Area of $$\triangle OPQ$$

The vertices are $$O=(0,0),\qquad P=\left(\sqrt5,\frac12\right),\qquad Q=\left(\frac{5\sqrt5}{4},0\right).$$

Using $$OQ$$ as the base and the $$y$$-coordinate of $$P$$ as the height,

$$\text{Area}=\frac12\times\text{base}\times\text{height}.$$

Thus,

$$\text{Area}=\frac12\times\frac{5\sqrt5}{4}\times\frac12.$$

Hence,

$$\text{Area}=\frac{5\sqrt5}{16}.$$

Therefore,

$$\boxed{\frac{5\sqrt5}{16}}.$$