A parabola whose vertex is at $(0,0)$ and opens upward is described by $\{(x,y)\in \mathbb R^2: 4py=x^2\}$ where the focus is at the point $(0,p)$ and the axis of the parabola is the $y$ axis. Therefore, a chord perpendicular to the axis of the parabola and passing through the focus is simply the line described by $\{(x,y)\in \mathbb R^2: y=p\}$.
This line intersects the parabola at $4p^2 = x^2$ or $x=\pm 2p$. The derivative of $y=\frac{x^2}{4p}$ is $y^{\prime} = \frac{x}{2p}$ which is $1$ at $x=2p$ and $-1$ at $x=-2p$. Since these slopes are the negative reciprocals of each other they are perpendicular.
Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$
I think this is correct.
and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
I don't think this is correct.
Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as
$$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$
So, in our case, we have
$$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$
Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines,
$$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$
Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.
As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.
Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).
In conclusion, the answer is $a^2\geqslant 8b^2$.
Best Answer
Let $A(at^2,2at)$ and $B(2bu,bu^2)$ be the points on $C_1:y^2=4ax$ and $C_2:x^2=4by$ respectively.
Equation of tangent of $C_1$ at $A$:
$$\frac{x}{-at^2}+\frac{y}{at}=1 \quad \cdots \cdots \: (1)$$
Equation of tangent of $C_2$ at $B$:
$$\frac{x}{bu}+\frac{y}{-bu^2}=1 \quad \cdots \cdots \: (2)$$ Comparing $(1)$ and $(2)$,
$$ \left \{ \begin{array}{rcl} -at^2 &=& bu \\ at &=& -bu^2 \\ \end{array} \right.$$
On solving,
$$t=\frac{1}{u}=-\sqrt[3]{\frac{b}{a}}$$
The common tangent is
$$-\frac{x}{\sqrt[3]{ab^2}}-\frac{y}{\sqrt[3]{a^2b}}=1$$
$$\fbox{$\sqrt[3]{a}\, x+\sqrt[3]{b} \, y+\sqrt[3]{a^2 b^2}=0$}$$