# What is the focus, vertex, and directrix of the parabola described by 16x^2=y?

Jul 24, 2017

Vertex is at $\left(0 , 0\right)$, directrix is $y = - \frac{1}{64}$ and focus is at $\left(0 , \frac{1}{64}\right)$.

#### Explanation:

$y = 16 {x}^{2} \mathmr{and} y = 16 {\left(x - 0\right)}^{2} + 0$ . Comparing with standard vertex form

of equation , y= a(x-h)^2 +k ; (h,k) being vertex , we find here

$h = 0 , k = 0 , a = 16$. So vertex is at $\left(0 , 0\right)$ . Vertex is at

equidistance from focus and directrix situated at opposite sides.

since $a > 0$ the parabola opens up . The distance of directrix from

vertex is $d = \frac{1}{4 | a |} = \frac{1}{4 \cdot 16} = \frac{1}{64}$ So directrix is $y = - \frac{1}{64}$ .

Focus is at $0 , \left(0 + \frac{1}{64}\right) \mathmr{and} \left(0 , \frac{1}{64}\right)$.

graph{16x^2 [-10, 10, -5, 5]} [Ans]

Jul 24, 2017

$\left(0 , \frac{1}{64}\right) , \left(0 , 0\right) , y = - \frac{1}{64}$

#### Explanation:

$\text{express the equation in standard form}$

$\text{that is } {x}^{2} = 4 p y$

$\Rightarrow {x}^{2} = \frac{1}{16} y$

$\text{this is the standard form of a parabola with the y-axis}$
$\text{as its principal axis and vertex at the origin}$

$\text{if 4p is positive graph opens up, if 4p is}$
$\text{negative the graph opens down}$

$\Rightarrow \textcolor{b l u e}{\text{vertex }} = \left(0 , 0\right)$

$\text{by comparison } 4 p = \frac{1}{16} \Rightarrow p = \frac{1}{64}$

$\text{focus } = \left(0 , p\right)$

$\Rightarrow \textcolor{red}{\text{focus }} = \left(0 , \frac{1}{64}\right)$

$\text{the directrix is a horizontal line below the origin}$

$\text{equation of directrix is } y = - p$

$\Rightarrow \textcolor{red}{\text{equation of directrix }} y = - \frac{1}{64}$