# What is the focus of the parabola x-4y^2+16y-19=0?

Jun 13, 2016

The coordinates of focus of the given parabola are $\left(\frac{49}{16} , 2\right) .$

#### Explanation:

$x - 4 {y}^{2} + 16 y - 19 = 0$
$\implies 4 {y}^{2} - 16 y + 16 = x - 3$
$\implies {y}^{2} - 4 y + 4 = \frac{x}{4} - \frac{3}{4}$
$\implies {\left(y - 2\right)}^{2} = 4 \cdot \frac{1}{16} \left(x - 3\right)$
This is a parabola along x-axis.
The general equation of a parabola along x-axis is ${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$,
where $\left(h , k\right)$ are coordinates of vertex and $a$ is the distance from vertex to the focus.

Comparing ${\left(y - 2\right)}^{2} = 4 \cdot \frac{1}{16} \left(x - 3\right)$ to the general equation, we get

$h = 3 , k = 2$ and $a = \frac{1}{16}$

$\implies$ $V e r t e x = \left(3 , 2\right)$

The coordinates of focus of a parabola along x-axis are given by $\left(h + a , k\right)$

$\implies F o c u s = \left(3 + \frac{1}{16} , 2\right) = \left(\frac{49}{16} , 2\right)$

Hence, the coordinates of focus of the given parabola are $\left(\frac{49}{16} , 2\right) .$