Question

How do you find the equation of a parabola with vertex at the origin and focus $\left( {0, - \dfrac{3}{2}} \right)$ ?

Answer 1

Here, we are given the vertex and the focus of a parabola and we need to calculate the equation of a parabola. It is given that the vertex is at the origin and so the vertex of the parabola is $\left( {0,0} \right)$.
The focus of a parabola is nothing but the fixed point that is used to draw the parabola and the vertex of the parabola is the point of intersection of the parabola and its axis.
Here, we need to find whether the given parabola is open rightward, open leftward, open upward or open downward. Also, we need to obtain the general equation of the parabola and we need to apply the given values.

Complete step-by-step answer:
It is given that the vertex is at the origin and focus is $\left( {0, - \dfrac{3}{2}} \right)$
That is the vertex of the parabola is $\left( {0,0} \right)$ and focus is $\left( {0, - \dfrac{3}{2}} \right)$
We shall visualize the given information by a graphical representation.

Hence, we can easily conclude that the given parabola is open downwards and the general equation of the parabola that is open downwards is ${x^2} = - 4ay$ …………….$\left( 1 \right)$
The given focus is on the form $f\left( {0, - a} \right)$and we shall compare it with $\left( {0, - \dfrac{3}{2}} \right)$
Hence, we got $a = \dfrac{3}{2}$
Now, we shall substitute $a = \dfrac{3}{2}$ in the equation $\left( 1 \right)$
Thus, we have ${x^2} = - 4ay$ $\Rightarrow {x^2} = - 4 \times \dfrac{3}{2}y$
$\Rightarrow {x^2} = - 2 \times 3y$
$\Rightarrow {x^2} = - 6y$
Hence, ${x^2} = - 6y$ is the required equation of the parabola.

Note: We got the desired equation of a parabola ${x^2} = - 6y$
Now, we shall do it.
{x^2} = - 6y$$$ \Rightarrow y = - \dfrac{1}{6}{x^2}$$ That is, $$y = - \dfrac{1}{6}{x^2}$$ is also the required equation of the parabola. We can stop up to {x^2} = - 6y$ and there is no need to solve it again.