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I solved this question but it is very lengthy.

Question: Find the equation of parabola with vertex $(2,-3)$ and focus $(0,5)$.

I'm adding here my answer which is lengthy enter image description here

cs89
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Raj Agarwal
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    Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. – José Carlos Santos Apr 17 '23 at 06:25
  • What means Vertex ? If you consider all parabols $y=a x^2+b x + c$ , you will find a solution. If you consider all paraboles $x = a y^2+b y +c$, you will find another solution, and if you consider all parabols (no restriction of the axis), you will find an infinity of solutions. – Lourrran Apr 17 '23 at 09:07
  • The vertex of a parabola is the point at the intersection of the parabola and its line of symmetry. – Raj Agarwal Apr 17 '23 at 11:38
  • Yes, we agree. Here, we know where this point is : (2,-3) ; On your picture, you decided that the line of symmetry should be horizontal. Why ? When I was at school, usually (or even always), the line of symmetry of a parabol was vertical. If you decide that the line of symmetry is horizontal (based on some external information), you have to search for an equation $x=ay^2 + bx+c$ ; and you shoud not have some terms in $xy$ in the final result. In both cases (line of symmetry horizontal or vertical), you should not have any $xy$ in all your formulas. Only $y = ax^2+bx+c$ or $x=ay^2+by+c$ – Lourrran Apr 17 '23 at 14:36
  • So this is general equation and how to find out this equation by these points that's the point mate. I tried to use distance formula which satisfies definition of parabola and got that equation. $y=ax^2+bx+c$ how to get this equation from these points that's the question. There is no need to remove x,y in the last equation which I wrote. That's correct. I want to know what is other idea do we have which is helpful to find out that equation quickly – Raj Agarwal Apr 17 '23 at 18:50
  • You have 3 unknowns (a,b,c). First equation is : when x=2, y=-3. So -3= 2^2a +2b+c; 2nd equation is : when x=0, y=5, so 5=0a+0b+c ; and 3rd equation is : (2,-3) is the vertex, so the line of symmetry is the line x=2 , so -b/2a=2 ; Are you ok with this equation n°3 ? The end is simple, when you have those 3 equations. – Lourrran Apr 17 '23 at 20:58
  • @Lourrran - The focus of a parabola also lies on its axis of symmetry. That means the axis of symmetry is the line passing though $(2,-3)$ and $(0, 5)$. This line is neither horizontal nor vertical. I.e., this parabola is not going to have either $y = ax^2 + bx +c$ nor $x = ay^2 + by + c$ as an equation. Raj Agarwal has apparently done the right thing (I can barely make out what is going on, so I cannot gauge how accurate it is - pictures of your handwriting are practically without any value). This why they ended up with an equation with $x^2, y^2,$ and $xy$ terms. – Paul Sinclair Apr 18 '23 at 03:02
  • @PaulSinclair thank you. Is there any idea to calculate the equation short method. – Raj Agarwal Apr 18 '23 at 05:02
  • Ok, understood. Focus is ... focus. A(2,-3) is THE vertex, and B(0,5) is THE focus. So we know the axis of this parabola, it is unique, and it is not vertical neither horizontal. The best way to solve this is probably to use a "change of reference". Build a new reference, so than AB is the vertical axis. You will find 'easily' the equation of the unique parabola in this new reference ($y= \frac{(x-a)^2}{2(b-k) }+\frac{b+k}{2}$ when the focus is $(a,b)$ and the directrix is $y=k$), and you will have simple formulas to go back to the original reference. – Lourrran Apr 18 '23 at 07:08
  • May I know how did you derive or get the formula? – Raj Agarwal Apr 18 '23 at 09:01
  • @RajAgarwal - no, I started writing up an approach, but when I was mostly done I looked back at your picture and realized it was almost exactly what you had done. I think the only improvement is I found the intersection of directrix and axis of symmetry slightly easier by reflecting the focus through the vertex. But only slightly. Unfortunately, Lourrran has not discussed the hard part - figuring out the change of coordinates, and then transforming the equation back. Once you add that in, I think you will find it harder/slower than what you did. – Paul Sinclair Apr 18 '23 at 11:27
  • https://fr.khanacademy.org/math/be-6eme-secondaire6h2/x4c2539af63a9a0cb:geometrie-dans-lespace-1/x4c2539af63a9a0cb:les-coniques/v/equation-for-parabola-from-focus-and-directrix This video explain how to get the formula (sound is in french, but formulas are international language) – Lourrran Apr 18 '23 at 11:39
  • I saw the whole video and it seems totally same to my method. Only they have solved it like a direct method. Can you explain me what is point of directrix here? Because in my method i got a equation for directrix but they have point like y=k – Raj Agarwal Apr 18 '23 at 16:57

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