Question

What are the vertex, focus and directrix of `y=x^2+10x+21`?

Answer 1

Vertex is `-5,-4)`, (focus is `(-5,-15/4)` and directrix is `4y+21=0`

Explanation:

Vertex form of equation is `y=a(x-h)^2+k` where `(h,k)` is vertex

The given equation is `y=x^2+10x+21`. It may be noted that the coefficient of `y` is `1` and that of `x` too is `1`. Hence, for converting the same, we have to make terms containing `x` a complete square i.e.

`y=x^2+10x+25-25+21` or

`y=(x+5)^2-4` or

`y=(x-(-5))^2-4`

Hence vertex is `(-5,-4)`

Standard form of parabola is `(x - h)^2=4p(y - k)`,

where focus is `(h,k+p)` and directrix `y=k-p`

As the given equation can be written as `(x-(-5))^2=4xx1/4(y-(-4))`, we have vertex `(h,k)` as `(-5,-4)` and

focus is `(-5,-15/4)` and directrix is `y=-5-1/4=-21/4` or `4y+21=0`