How to get the equation of tangent of $x^2=4ay$ from the equation of tangent of $y^2=4ax$?

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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.

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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 $.

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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 $y$.