Label the vertex point, tangent point, cone max radius point and center taken anticlockwise as $ V,T,B , O$ respectively .
It is convenient to take height $ h = VO $.
Verify the distances from trigonometry of right angled triangles.
$ VT = \sqrt {(h^2-r^2)}$
For the right angled triangle $ OVTB$ .. triangles $ OTV, VTO $ are similar; so $ TO^2 = TV \cdot TB $
$ TB = \dfrac{r^2}{\sqrt {(h^2-r^2)} }$
$ BO =\dfrac{h\cdot r}{\sqrt {(h^2-r^2)}} $
$ BT =\dfrac{ r^2}{\sqrt {(h^2-r^2)}} $
Slant area = $ \pi\cdot BO \cdot VB = \pi\cdot r\cdot h^3/ (h^2-r^2) $
Differentiate and simplify to get
$ h/r = \csc \alpha = \csc TVO = \sqrt 3 $
$ \alpha = ~ 35.25 \; degrees $
Volume of cone $ V = \pi\cdot BO^2 \cdot h/3 $
leads to the desired result.
Best Answer
Let's apply the sine rule to the upper triangle on the right: if $p_1$ is its horizontal (blue) side we have $$ {p_1\over\sin(\pi/2+\Omega)}={R\tan\Omega\over\sin(\pi/2-\Omega-\alpha)}, \quad\hbox{whence:}\quad p_1=R\tan\Omega{\cos\Omega\over\cos(\alpha+\Omega)}. $$ In an analogous way, by applying the sine rule to the upper triangle on the left, its horizontal (blue) side turns out to be: $$ p_2=R\tan\Omega{\cos\Omega\over\cos(\alpha-\Omega)}. $$ The major axis $2a$ of the ellipse is then $2a=p_1+p_2$.
To find semi-major axis $b$ we can substitute $x=(p_1-p_2)/2$ and $y=R\tan\Omega$ into the equation $$ {x^2\over a^2}+{y^2\over b^2}=1, $$ which gives: $$ b={R\tan\Omega(p_1+p_2)\over2\sqrt{p_1p_2}}. $$
From $a$ and $b$ you can compute the area of the ellipse.