Question

What are the vertex, focus and directrix of `y=4x^2 + 5x + 7`?

Answer 1

Given equation:

`y=4x^2+5x+7`

`y=4(x^2+5/4x)+7`

`y=4(x^2+5/4x+25/64)-25/64+7`

`y=4(x+5/8)^2+423/64`

`(x+5/8)^2=1/4(y-423/64)`

Comparing above equation with the standard form of parabola `X^2=4aY` we get

`X=x+5/8, \ Y=y-423/64, a=1/16`

Vertex of Parabola

`X=0, Y=0`

`x+5/8=0, y-423/64=0`

`x=-5/8, y=423/64`

`(-5/8, 423/64)`

Focus of parabola

`X=0, Y=a`

`x+5/8=0, y-423/64=1/16`

`x=-5/8, y=427/64`

`(-5/8, 427/64)`

Directrix of parabola

`Y=-a`

`y-423/64=-1/16`

`y=419/64`