Question

How do you find the standard form of the equation for the parabola with vertex (4,0) and passing through the point (1,2)?

Answer 1

Equation is `y=2/9x^2-16/9x+32/9` or `x=-3/4y^2+4`

Explanation:

With vertex at `(h,k)`, we can have two types of parabolas - (i) of regular vertical type `y=a(x-h)^2+k` and (ii) of horizontal type `x=a(y-k)^2+h`. In standard form it is `y=ax^2+bx+c` or `x=ay^2+by+c`

If it is regular type, if vertex is `(4,0)`, our equation is

`y=a(x-4)^2+0` and as it passes through `(1,2)`, we have

`2=a(1-4)^2` or `9a=2` i.e. `a=2/9` and

equation is `y=2/9(x-4)^2` or `9y=2(x-4)^2`

or in standard form equation is `y=2/9x^2-16/9x+32/9`

If parabola is horizontal, if vertex is `(4,0)`, our equation is

`x=a(y-0)^2+4` and as it passes through `(1,2)`, we have

`1=a*2^2+4` or `4a=-3` i.e. `a=-3/4` and

equation is `x=-3/4y^2+4` or `4x+3y^2=16`

and in standard form equation is `x=-3/4y^2+4`
graph{(4x+3y^2-16)(2(x-4)^2-9y)=0 [-10, 10, -5, 5]}