Tangent at a Point

The point of intersection of the two ellipses satisfies

$$x^2+4y^2=1$$ and $$4x^2+y^2=1$$

Subtracting,

$$3x^2-3y^2=0$$

$$x^2=y^2$$

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

$$x=y$$

Substituting into

$$x^2+4y^2=1$$ gives $$5x^2=1$$

$$x=y=\frac1{\sqrt5}$$

Hence,

$$P=\left(\frac1{\sqrt5},\frac1{\sqrt5}\right)$$

For the ellipse

$$x^2+4y^2=1$$

differentiating implicitly,

$$2x+8y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{4y}$$

At $$P$$,

$$m_1=-\frac14$$

For the ellipse

$$4x^2+y^2=1$$

differentiating implicitly,

$$8x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{4x}{y}$$

At $$P$$,

$$m_2=-4$$

The angle between the tangents is given by

$$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$$

Substituting,

$$\tan\theta=\left|\frac{-4+\frac14}{1+1}\right|$$

$$=\frac{\frac{15}{4}}{2}$$

$$=\frac{15}{8}$$

Therefore,

$$4\tan\theta=4\cdot\frac{15}{8}$$

$$=\frac{15}{2}$$

Hence,

$$\boxed{\frac{15}{2}}$$

The ellipse is

$$\frac{x^2}{9}+\frac{y^2}{4}=1$$

and the external point is

$$P(3,4)$$

Let the points of contact of the tangents be

$$A(x_1,y_1)$$

and

$$B(x_2,y_2)$$

First find the chord of contact.

For the ellipse

$$\frac{x^2}{9}+\frac{y^2}{4}=1$$

the chord of contact from

$$P(3,4)$$

is

$$\frac{3x}{9}+\frac{4y}{4}=1$$

$$\frac{x}{3}+y=1$$

$$x+3y-3=0$$

Thus

$$AB:x+3y-3=0$$

The slope of

$$AB$$

is

$$-\frac13$$

Hence the altitude from

$$P$$

has slope

$$3$$

Therefore,

$$y-4=3(x-3)$$

$$y=3x-5$$

Now find the points of contact.

The tangent to the ellipse at

$$(x_1,y_1)$$

is

$$\frac{xx_1}{9}+\frac{yy_1}{4}=1$$

Since it passes through

$$P(3,4)$$

we get

$$\frac{x_1}{3}+y_1=1$$

Also,

$$\frac{x_1^2}{9}+\frac{y_1^2}{4}=1$$

Substituting

$$y_1=1-\frac{x_1}{3}$$

into the ellipse:

$$\frac{x_1^2}{9}+\frac{1}{4}\left(1-\frac{x_1}{3}\right)^2=1$$

Multiplying by

$$36$$

gives

$$4x_1^2+9-6x_1+x_1^2=36$$

$$5x_1^2-6x_1-27=0$$

$$x_1=\frac{6\pm24}{10}$$

Hence

$$x_1=3,\quad x_1=-\frac95$$

Corresponding values of

$$y_1$$

are

$$0,\quad \frac85$$

Therefore

$$A=(3,0)$$

and

$$B=\left(-\frac95,\frac85\right)$$

Now find two altitudes.

Slope of

$$PB$$

is

$$\frac{\frac85-4}{-\frac95-3}$$

$$=\frac{-\frac{12}{5}}{-\frac{24}{5}}$$

$$=\frac12$$

Therefore the altitude through

$$A$$

has slope

$$-2$$

Its equation is

$$y=-2(x-3)$$

$$y=-2x+6$$

The altitude through

$$P$$

was

$$y=3x-5$$

The orthocenter is the intersection of these two lines.

Thus,

$$3x-5=-2x+6$$

$$5x=11$$

$$x=\frac{11}{5}$$

Substituting,

$$y=3\left(\frac{11}{5}\right)-5$$

$$=\frac{33}{5}-\frac{25}{5}$$

$$=\frac{8}{5}$$

Hence the orthocenter is

$$\left(\frac{11}{5},\frac{8}{5}\right)$$

Final Answer :

$$\left(\frac{11}{5},\frac{8}{5}\right)$$

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,

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