What is the focus of the parabola $x + 5 {(y - 3)}^{2} = 6$ $x + 5(y - 3)^2 = 6$ ?

Question

1788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-23T02:24:20

$(\frac{119}{20}, 3)$ $(119/20, 3)$

Write in vertex form for a left or right opening parabola, $x = a {(y - k)}^{2} + h$ $x = a(y - k)^2 + h$ :

#x = -5(y - 3)^2 + 6

The directed focal length, f, for all parabolas is:

$f = \frac{1}{4 a}$ $f = 1/(4a)$

Substitute -5 for a:

$f = \frac{1}{4 (- 5)}$ $f = 1/(4(-5))$

$f = - \frac{1}{20}$ $f = -1/20$

The general form for the focus of a left or right opening parabola is:

$(h + f, k)$ $(h + f, k)$

Substitute 6 for h, 3 for k, and $- \frac{1}{20}$ $-1/20$ for f:

$(6 + - \frac{1}{20}, 3)$ $(6 + -1/20, 3)$

$(\frac{119}{20}, 3)$ $(119/20, 3)$

What is the focus of the parabola x + 5(y - 3)^2 = 6x+5(y−3)2=6x + 5(y - 3)^2 = 6?