Question

Find the equation of a parabola, whose vertex is at `(-3,2)` and passes through `(4,7)`?

Answer 1

`5x^2+30x-49y+143=0` or `7y^2-25x-28y-47=0`

Explanation:

There could be two type of parabolas with vertex as `(-3,2)`

A `(y-2)=a(x+3)^2` If this passes through `(4,7)`, we have

`(7-2)=a(4+3)^2` i.e. `a=5/49` and equation of parabola is

`(y-2)=5/49(x+3)^2` i.e. `49y-98=5x^2+30x+45` or

`5x^2+30x-49y+143=0`

B `(x+3)=a(y-2)^2` If this passes through `(4,7)`, we have

`(4+3)=a(7-2)^2` i.e. `a=7/25` and equation of parabola is

`(x+3)=7/25(y-2)^2` i.e. `25x+75=7y^2-28y+28` or

`7y^2-25x-28y-47=0`

graph{(5x^2+30x-49y+143)(7y^2-25x-28y-47)=0 [-11, 9, -1.36, 8.64]}