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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?

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    $\begingroup$ Since the parabola is upside down, the focus is at $(0,-a)$ $\endgroup$ – David Quinn Feb 19 '16 at 16:58
  • $\begingroup$ You seem to be assuming the parabola passes through $(0,0)$ $\endgroup$ – Henry Feb 19 '16 at 16:59
  • $\begingroup$ There is no reason to assume it opens downward. $y=A x^2+B$, where $-3=4 A+B$, and $A\ne 0$, is a parabola thru $(2,-3)$. $\endgroup$ – DanielWainfleet Feb 19 '16 at 17:57
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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.

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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: $$ y=ax^2-4a-3 $$

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  • $\begingroup$ If a problem statement says something like "The focus is at (6.0) and directrix is along x = -6), would that be sufficient to conclude that the equation of the parabola is $y^2 = 24x$ then? $\endgroup$ – Aayush Agrawal Feb 19 '16 at 17:47
  • $\begingroup$ Yes it is. Note these conditions also fix that the parabola is right-wards open. $\endgroup$ – Emilio Novati Feb 19 '16 at 17:51
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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).$

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