Why is the focus of the parabola not within the parabola in the following result?

Question

So i'm going through my book and try to solve the following question:

Find the equation of the parabola which is symmetric about the y axis and passed through the point (2,-3).

Since it passes through (2,-3), we can assume that the parabola opens downwards, and hence use the equation $x^2 = -4ay$.

Plugging in the values though you'd get $4 = -4(-3)a$ or $a = \frac13$

But this implies that the focus is at $(0, \frac13)$ which is clearly not in the parabola. How is this possible/where did i go wrong?

Answer 1

You are confusing yourself here! If a parabola has equation $x^2=4ay$, then its focus is at $(0,a)$. But you have given your parabola the equation $x^2=-4ay$, so its focus is at $(0,-a)$.

Simpler, I would say, to stick with $x^2=4ay$ and let $a$ be negative.

Answer 2

The parabola does not go through origin. It is of the form

$$ x^2 = b^2 - 4 a y $$

Depending on value of $b$ chosen there can be an infinite number of parabolas.They all pass through $ (2,-3),(-2,-3).$

Answer 3

You are wrong when you say: ''Since it passes through (2,-3), we can assume that the parabola opens downwards''. Really, if we know only the symmetry axis and a point we cannot say if the parabola is opend upwards or downwards.

You can only say that the equation of all the parabolas that are symmetric with respect to the $y$ axis have an equation of the form: $$ y=ax^2+c $$ and , if the parabola passes through the point $(2,-3)$, substituting the coordinates we find $c=-3-4a$, so the equation has the form: $$ y=ax^2-4a-3 $$