I'm assuming you don't want any calculus involved here (too bad!), so let $\,(a,b)\,,\,(c,d)\,$ be the points on the parabola through which pass two tangent lines to it that are perpendicular:
$$\text{First tangent: we need to solve the system}\;\;\;\;y^2=4px\;\;,\;\;y-b=m(x-a)\Longrightarrow$$
$$(m(x-a)+b)^2=4px$$
$$\text{Second tangent: we need to solve the sytem}\;\;\;\;y^2=4px\;\;,\;\;y-d=-\frac{1}{m}(x-c)\Longrightarrow$$
$$\left(-\frac{1}{m}(x-c)+d\right)^2=4px$$
Of course, solving the above take into account that
$$b^2=4pa\;\;,\;\;d^2=4pc$$
since both points were chosen to be on the parabola.
Also, remember that two straight lines (none of which is horizontal/vertical) with slopes $\,m_1\,,\,m_2\,$ are perpendicular iff $\,m_1m_2=-1\,$, and this the reason we took the second tangent's slope to be $\,-1/m\,$
Expanding upon my comment, and following the same basic strategy as in this answer, we can get the equation mechanically using the determinant for the five-point conic through $P=(P_x,P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$:
$$\left|\begin{array}{c,c,c,c,c,c}
x^2 & y^2 & x y & x & y & 1 \\
P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\
Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\
R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\
S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\
T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\
\end{array}\right| = 0 \tag{$\star$}$$
It happens to be convenient to have the origin at the midpoint of the given points, so let us take
$$P = (d\cos\theta,d\sin\theta) \qquad Q = (-d\cos\theta, -d\sin\theta) \tag{1}$$
Two more points come from infinitesimally-displacing $P$ and $Q$ along their tangent lines (in directions, say, $\phi$ and $\psi$, respectively):
$$R = P + (p\cos\phi, p\sin\phi) \qquad S = Q + (q\cos\psi,q\sin\psi) \tag{2}$$
for "very small" $p$ and $q$. The fifth point is provided by the specific geometry of the parabola; if $M$ is the midpoint of $P$ and $Q$, and $N$ is the point where the tangent lines at $P$ and $Q$ meet, then the midpoint of $M$ and $N$ lies on the parabola. Thus, we can take
$$T = \frac{d}{2\sin(\phi-\psi)}\left(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta), \sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta)\right) \tag{3}$$
Substituting into $(\star)$, and letting a computer algebra system crunch some symbols, we factor-out and cancel a $p$ and $q$, then set the remaining $p$s and $q$s to zero. Canceling factors of $d^2 \csc^2(\phi-\psi) \sin(\phi-\theta)\sin(\psi - \theta)$, we have
$$\begin{align}
0 &= \phantom{2}x^2 (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))^2 \\
&+ \phantom{2}y^2 (\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta))^2 \\
&-2x y (\sin\phi\sin(\psi-\theta)+\sin\psi\sin(\phi-\theta))(\cos\phi\sin(\psi-\theta)+\cos\psi\sin(\phi-\theta)) \\
&-4 x d \sin\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\
&+4 y d \cos\theta \sin(\phi-\psi) \sin(\phi-\theta) \sin(\psi-\theta) \\
&-4 d^2 \sin^2(\phi-\theta)\sin^2(\psi-\theta)
\end{align} \tag{$\star\star$}$$
Best Answer
From this post, we can write the equation of a parabola as $$d_{\text{Axis}}^2=4a\cdot d_{\text{Tangent at vertex}}.$$ Here $d_{\text{line}}$ represents the algebraic distance of any arbitrary point on the parabola from the said line. $**$(More intuitive explanation below).
Here, tangent at vertex is $y=kx$.
The axis (straight line through the vertex (that you referred to as the base) and perpendicular to tangent at vertex) is $y+\dfrac xk=y_1+\dfrac{x_1}{k}$.
Thus the equation of parabola is $$\bigg(\frac{ky+x-(ky_1+x_1)}{\sqrt{1+k^2}}\bigg)^2=4a\cdot\dfrac{y-kx}{\sqrt{1+k^2}}$$
Put $(x_2, y_2)$ into the equation and find $a$. Done :)
$**$ Intuitive explanation: Consider the standard parabola $y^2=4ax$. Here $y$= Distance from X axis ($\equiv$ Axis) and $x$= Distance from Y axis ($\equiv $ Tangent at vertex)