Question

What is the equation of the parabola that has a vertex at `(-2, 3)` and passes through point `(13, 0)`?

Answer 1

equation of parabola can be expressed as, `y=a(x-h)^2 +k` where, `(h,k)` is the coordinate of vertex and `a` is a constant.

Given,`(h,k)=(-2,3)` and the parabola passes through `(13,0)`,

So,putting the values we get,

`0=a(13-(-2))^2 +3`

or, `a=-3/225`

So,the equation becomes, `y=-3/225(x+2)^2 +3` graph{y=(-3/225)(x+2)^2 +3 [-80, 80, -40, 40]}

Answer 2

`y=-1/75(x+2)^2+3`

or `x=5/3(y-3)^2-2`

Explanation:

We can make two types of parabolas, one vertical and other horizontal. The equation of vertical parabola, whose vertex is `(-2,3)` is

`y=a(x+2)^2+3` and as it passes through `(13,0)`, we have

`0=a(13+2)^2+3` or `a=(-3)/15^2=-3/225=-1/75`

and hence equation is `y=-1/75(x+2)^2+3`

The curve appears as follows:

graph{(y+1/75(x+2)^2-3)((x+2)^2+(y-3)^2-0.08)=0 [-20, 20, -10, 10]}

The equation of horizontal parabola, whose vertex is `(-2,3)` is

`x=a(y-3)^2-2` and as it passes through `(13,0)`, we have

`13=a(0-3)^2-2` or `a=(13+2)/3^2=15/9=5/3`

and hence equation is `x=5/3(y-3)^2-2`

The curve appears as follows:

graph{(x-5/3(y-3)^2+2)((x+2)^2+(y-3)^2-0.08)=0 [-20, 20, -10, 10]}