How do you write the equation using polar coordinates given ${x^2} = 4y$?

Hint: In this question, we need to express the given equation in terms of polar coordinates. The given equation is in the form of Cartesian coordinate. Here we will simply substitute the value of the variable x and y as, $x = r\cos \theta $ and $y = r\sin \theta $ in the given equation and solve it. We find out the value for $r$ and simplify the problem given. Then after solving it we will write it in the simplified form which will be the polar coordinate form of the given equation.

Complete step by step solution:
Given the equation of the form ${x^2} = 4y$ …… (1)
We are asked to represent the above equation (1) in terms of polar coordinates.
The given equation is in the form of Cartesian coordinate.
 To convert the given equation into the polar form we will make substitution for the variable x and y.
We substitute $x = r\cos \theta $ and $y = r\sin \theta $, where $r = \sqrt {{x^2} + {y^2}} $
Substituting the values of x and y in the equation (1), we get,
${(r\cos \theta )^2} = 4 \cdot r\sin \theta $
Now we will simply open the parenthesis and square the terms in the parenthesis.
Therefore, we get,
$ \Rightarrow {r^2}{\cos ^2}\theta = 4 \cdot r\sin \theta $
Now dividing by $r\sin \theta $ in the R.H.S. and L.H.S. we get,
$ \Rightarrow \dfrac{{{r^2}{{\cos }^2}\theta }}{{r\sin \theta }} = \dfrac{4}{{r\sin \theta }} \cdot r\sin \theta \cdot $
Now cancelling the terms in numerator and denominator we get,
$ \Rightarrow \dfrac{{r{{\cos }^2}\theta }}{{\sin \theta }} = 4$
Taking $\sin \theta $ to the other side we get,
$ \Rightarrow r{\cos ^2}\theta = 4\sin \theta $
Now we will take the term ${\cos ^2}\theta $ to the other side of the equation we get,
$ \Rightarrow r = \dfrac{{4\sin \theta }}{{{{\cos }^2}\theta }}$
This also can be written as,
$ \Rightarrow r = \dfrac{{4\sin \theta }}{{\cos \theta \cdot \cos \theta }}$
$ \Rightarrow r = 4 \cdot \dfrac{{\sin \theta }}{{\cos \theta }} \cdot \dfrac{1}{{\cos \theta }}$
We know the trigonometric functions, $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ and $\dfrac{1}{{\cos \theta }} = \sec \theta $.
Hence we get,
$ \Rightarrow r = 4\tan \theta \sec \theta $

Hence polar coordinate representation of the equation ${x^2} = 4y$ is given by $r = 4\tan \theta \sec \theta $.

Note: Here we have to remember that the ratio of the $\sin \theta $ and $\cos \theta $ is equal to the $\tan \theta $.
Also the reciprocal of the cosine function is equal to secant function.
i.e. $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ and $\dfrac{1}{{\cos \theta }} = \sec \theta $
We don’t have to confuse the polar coordinate system with the normal rectangular coordinate system. Polar coordinate system is the system in which the coordinates of a point is represented by the distance of that point from a reference point and by the angle from the reference plane.
i.e. we substitute $x = r\cos \theta $ and $y = r\sin \theta $ in the place of x and y.