2. A conic section with centre at $$ (\sqrt{2},-1) $$ has the value of eccentricity given by $$ e=\frac{3 \sqrt{2}}{2} $$. Given that the focal axis is horizontal and that the distance between foci and directrices is seven thirds, (a) write down the conic section's equation (b) write down the equations of directrices.

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We are told that the conic has a centre at 
$$
(\sqrt{2}, -1),
$$
eccentricity 
$$
e=\frac{3\sqrt{2}}{2},
$$
and that its focal axis is horizontal. (Since $$ e>1 $$ the conic is a hyperbola.) Its foci are then located at 
$$
\left(\sqrt{2}\pm c,\,-1\right),
$$
where $$ c=e a $$ and $$ a>0 $$ is the distance from the centre to a vertex. Also, the distance from the centre to a directrix is 
$$
\frac{a}{e},
$$
so that the two directrices (vertical lines) are given by 
$$
x=\sqrt{2}\pm \frac{a}{e}.
$$

We are further given that the distance from each focus to its corresponding directrix is 
$$
\frac{7}{3}.
$$
Since the focus is further from the centre than the directrix, the distance between a focus and the corresponding directrix is
$$
\text{(distance from focus to centre)} - \text{(distance from directrix to centre)} = c - \frac{a}{e}.
$$
Thus, we have
$$
c-\frac{a}{e}=\frac{7}{3}.
$$
But $$ c=e a $$. Hence,
$$
e a-\frac{a}{e}=\frac{7}{3}.
$$
We factor out $$ a $$:
$$
a\left(e-\frac{1}{e}\right)=\frac{7}{3}.
$$
This gives
$$
a=\frac{7}{3\left(e-\frac{1}{e}\right)}.
$$
It is often more convenient to write the difference in a single fraction:
$$
e-\frac{1}{e}=\frac{e^2-1}{e},
$$
so that
$$
a=\frac{7}{3\cdot \frac{e^2-1}{e}}=\frac{7e}{3(e^2-1)}.
$$

Now, substitute $$ e=\frac{3\sqrt{2}}{2} $$. First find $$ e^2 $$:
$$
e^2=\left(\frac{3\sqrt{2}}{2}\right)^2=\frac{9\cdot 2}{4}=\frac{18}{4}=\frac{9}{2}.
$$
Thus,
$$
e^2-1=\frac{9}{2}-1=\frac{9-2}{2}=\frac{7}{2}.
$$
Then,
$$
a=\frac{7e}{3\cdot \frac{7}{2}}=\frac{7e}{\frac{21}{2}}=\frac{7e\cdot2}{21}=\frac{14e}{21}=\frac{2e}{3}.
$$
Substitute back $$ e=\frac{3\sqrt{2}}{2} $$:
$$
a=\frac{2}{3}\cdot\frac{3\sqrt{2}}{2}=\sqrt{2}.
$$

We now have:
$$
a=\sqrt{2}\quad \text{and} \quad c=e\,a=\frac{3\sqrt{2}}{2}\cdot\sqrt{2}=\frac{3\cdot2}{2}=3.
$$

For a hyperbola with horizontal transverse axis and centre $$(h,k)=(\sqrt{2},-1)$$ the standard equation is
$$
\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,
$$
and the relation among the parameters is
$$
c^2=a^2+b^2.
$$
Thus,
$$
9=2+b^2\quad \Longrightarrow \quad b^2=7.
$$

The hyperbola’s equation is therefore
$$
\frac{(x-\sqrt{2})^2}{2}-\frac{(y+1)^2}{7}=1.
$$

The directrices are vertical lines at distance $$ \frac{a}{e} $$ from the centre. We already have
$$
a=\sqrt{2},\quad e=\frac{3\sqrt{2}}{2}.
$$
So,
$$
\frac{a}{e}=\frac{\sqrt{2}}{\frac{3\sqrt{2}}{2}}=\frac{\sqrt{2}\times2}{3\sqrt{2}}=\frac{2}{3}.
$$
Thus, the directrices are given by
$$
x=\sqrt{2}\pm \frac{2}{3}.
$$

(a)
$$
\frac{(x-\sqrt{2})^2}{2}-\frac{(y+1)^2}{7}=1.
$$

(b)
$$
x=\sqrt{2}-\frac{2}{3}\quad \text{and}\quad x=\sqrt{2}+\frac{2}{3}.
$$
(a) The equation of the hyperbola is: $$ \frac{(x-\sqrt{2})^2}{2} - \frac{(y+1)^2}{7} = 1 $$ (b) The equations of the directrices are: $$ x = \sqrt{2} - \frac{2}{3} \quad \text{and} \quad x = \sqrt{2} + \frac{2}{3} $$

A conic section with centre at has the value of eccentricity given by . Given that
the focal axis is horizontal and that the distance between foci and directrices is seven thirds,
(a) write down the conic section’s equation
(b) write down the equations of directrices.

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A conic section with centre at has the value of eccentricity given by . Given that the focal axis is horizontal and that the distance between foci and directrices is seven thirds, (a) write down the conic section’s equation (b) write down the equations of directrices.