The equation of the normal to the parabola $y^2=4ax$ is $y=mx-2am-am^3$ and the parametric is $y+tx=2at+at^3$. What would be these equations be if the parabola is $x^2=4ay$ ? I was taught that when $x$ and $y$ for a parabola, ellipse or hyperbola is interchanged, for the corresponding equations of normal and tangent we swap the places of $ x$ and $y$ and put $m \ as \ \frac{1}{m}$. But today I came across the normal equation of parabola as $x=my-2am-am^3$ and $x+yt=2at+at^3$ in solutions where there is no change in $t$ and $m$.
The same doubt as above where x and y is swapped except for a tangent. Tangent of $y^2=4ax$ is $y=mx+\frac{a}{m}$ and parametric is $ty=x+at^2$
In such scenario consider $m$ as constant not as slope. So if tangent of $y^2 = 4ax$ is $y=mx + \frac{a}{m}$, for the tangent of $x^2 = 4ay$ just interchange $x$ and $y$ i.e. $x = my + \frac{a}{m}$.
Now if you wish to know slope you can find by coefficient comparison.
Same for normal: $y^2 = 4ax\implies y=mx - 2am - am^3$. The tangent of $x^2 = 4ay\implies$ just interchange $x$ and $y$ i.e. $x = my - 2am - am^3$.
In terms of parametric $t$:
$y^2 = 4ax\implies y=x/t + at$. The tangent of $x^2 = 4ay\implies$ just interchange $x$ and $y$ i.e. $x = y/t + at$.
Same for normal: $y^2 = 4ax\implies y=-tx + 2at + at^3$. The tangent of $x^2 = 4ay\implies$ just interchange $x$ and $y$ i.e. $x = -ty + 2at + at^3$.
This trick can also be useful if parabola is shifted i.e. for $(y-k)^2 = 4a(x-h)\implies (y-k)=m(x-h) + a/m$. The tangent of $(x-h)^2 = 4a(y-k)\implies (x-h) = m(y-k) + a/m$.