How to get the equation of tangent of $x^2=4ay$ from the equation of tangent of $y^2=4ax$? We know that the equation of tangent of $y^2=4ax$ is $y=mx+\frac{a}{m}$.
Now on replacing $x$ to $y$ and $m$ to $\frac{1}{m}$ in the previous equation, we get, $y=mx-am^2$ which is the tangent of $x^2=4ay$. Why does that happen?
I just can't understand the logic behind this transformation. I somewhat get the feel of replacing $x$ to $y$. But that would just rotate the axis.
 A: When swapping $x$ and $y$, for both the function and the tangent line, you're basically finding the reflection of the original parabola, and the tangent line, about the line $y = x$.
Swapping $x$ and $y$ for the parabola results in $ x^2 = 4 a y $
And swapping $x$ and $y$ for its tangent results in $ x = m y + \dfrac{a}{m} $ which, when re-arranged, becomes $ y = \dfrac{1}{m} x - \dfrac{a}{m^2} $
If you define $m' = \dfrac{1}{m} $ then the reflected tangent equation is $ y = m' x - a m'^2 $.
A: My approach is to actually observe the relation between $y^2=4ax$ and $x^2=4ay$. They both are basically the same curves but just rotated the axis by $90^0$. We'll do the same here.
Here, the tangent equation of $y^2=4ax$ is $y=mx+\frac{a}{m}$. Now, we need to rotate the co-ordinate system in clockwise direction by $90^0$.
Observe that we are not rotating the graph here but the coordinate system. Thus, you'll get the clockwise rotation. To do that we'll use the rotation of axes formulae here.
$$
x=Xcos\theta-Ysin\theta ~ ~
y=Xsin\theta+Ycos\theta $$
And here, $\theta=-90^0$ as here angle is measured in the anticlockwise direction. Thus, we'll get $x=Y, ~ y=-X$. This means, New tangent equation will be
$$ X+mY+\frac{a}{m}=0$$
Now, as we have changed the coordinate system, the definition of slope with also change. Thus, replace $m$ with $\frac{-1}{m}$. Now, you'll get a neat equation for the tangent.
$$ mX-Y-am^2=0 \\ \\ Y=mX-am^2$$
You can just ignore those Capitals and think of them as normal $x$ and $y$.
