The number of those tangents to the curve $y^2-2x^3-4y+8=0$ which pass through the point $\left(1,2\right)$ is

Let $P\left(\alpha ,\beta \right)$ be any point on the curve

$y^2-2x^3-4y+8=0$

Now, $2y\frac{dy}{dx}-6x^2-4\frac{dy}{dx}=0$

$\Rightarrow \left(2y-4\right)\frac{dy}{dx}-6x^2=0$

$\Rightarrow \frac{dy}{dx}=\frac{6x^2}{\left(2y-4\right)}$

Thus, slope $=m={\left(\frac{dy}{dx}\right)}_P=\frac{6{\alpha }^2}{\left(2\beta -4\right)}$
Now, the equation of the tangent at $P$ is
$y-\beta =\frac{6{\alpha }^2}{\left(2\beta -4\right)}\left(x-\alpha \right)$
which is passing through $\left(1,2\right)$
$\Rightarrow 2-\beta =\frac{6{\alpha }^2}{\left(2\beta -4\right)}\left(1-\alpha \right)$

$\Rightarrow 2-\beta =\frac{3{\alpha }^2}{\left(\beta -2\right)}\left(1-\alpha \right)$

$\Rightarrow -{\left(\beta -2\right)}^2=3{\alpha }^2\left(1-\alpha \right)$

$\Rightarrow 3{\alpha }^3-3{\alpha }^2-{\beta }^2+4\beta -4=0 ...\left(\mathrm{i}\right)$

Also, $P\left(\alpha ,\beta \right)$ lies on $y^2-2x^3-4y+8=0$

So, ${\beta }^2-2{\alpha }^3-4\beta +8=0 ...\left(\mathrm{ii}\right)$

From (i) and (ii), we get,

$\Rightarrow -{\alpha }^3+3{\alpha }^2-4=0$

$\Rightarrow {\alpha }^3-3{\alpha }^2+4=0$

$\Rightarrow {\alpha }^3+{\alpha }^2-4{\alpha }^2-4\alpha +4\alpha +4=0$

$\Rightarrow {\alpha }^2\left(\alpha +1\right)-4\alpha \left(\alpha +1\right)+4\left(\alpha +1\right)=0$

$\Rightarrow \left(\alpha +1\right)\left({\alpha }^2-4\alpha +4\right)=0$

$\Rightarrow \left(\alpha +1\right){\left(\alpha -2\right)}^2=0$

when $\alpha =-1$, then ${\beta }^2+2-4\beta +8=0$

Thus, $\beta$ is imaginary.

when $\alpha =2$, then ${\beta }^2-16-4\beta +8=0$

$\Rightarrow {\beta }^2-4\beta -8=0$

$\Rightarrow {\left(\beta -2\right)}^2=12={\left(2\sqrt{3}\right)}^2$

$\Rightarrow \left(\beta -2\right)=\pm \left(2\sqrt{3}\right)$

$\Rightarrow \beta =2\pm 2\sqrt{3}$

Hence, the point of contacts are
$\left(2,2+2\sqrt{3}\right)$ and $\left(2,2-2\sqrt{3}\right)$
Slope of the tangents are $2\sqrt{3},-2\sqrt{3}$
Hence, the equations of tangents are
$y-\left(2+2\sqrt{3}\right)=2\sqrt{3}\left(x-2\right)$
and
$y-\left(2-2\sqrt{3}\right)=-2\sqrt{3}\left(x-2\right)$

Hence two tangent are possible.