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, $$\text{Focus}=\left(2\cdot\frac{\sqrt5}{2},0\right)=(\sqrt5,0).$$
---
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,
$$\frac{4x^2}{25}+\frac{4y^2}{5}=1.$$
---
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,
$$P=\left(\sqrt5,\frac12\right).$$
---
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,
$$Q=\left(\frac{5\sqrt5}{4},0\right).$$
---
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,
$$\boxed{\frac{5\sqrt5}{16}}.$$
91- 6303239042
