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What is the focus of the parabola $y = 10x^2 + 3$ ? $\newline$ Simplify any fractions. $\newline$ (,)

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Find the focus or directrix of a parabola

Full solution

Q. What is the focus of the parabola

y = 10x^2 + 3

?

\newline

Simplify any fractions.

\newline

(______,______)

Orientation: Vertical: We have: $y = 10x^2 + 3$ $\newline$ Is the parabola horizontal or vertical? $\newline$ $y = 10x^2 + 3$ is in the form of $y = a(x - h)^2 + k$ . $\newline$ Orientation: Vertical
Identify Values: $y = 10x^2 + 3$ $\newline$ Identify the values of $a$ , $h$ and $k$ using the vertex form. $\newline$ Compare $y = a(x - h)^2 + k$ and $y = 10(x - 0)^2 + 3$ . $\newline$ $a = 10$ $\newline$ $h = 0$ $\newline$ $k = 3$
Find $p$ Value: We found: $\newline$ $a = 10$ $\newline$ What is the value of $p$ ? $\newline$ Substitute $10$ for $a$ in $p = \frac{1}{4a}$ . $\newline$ $p = \frac{1}{4\times 10}$ $\newline$ $p = \frac{1}{40}$
Focus of Parabola: We know: $\newline$ $h = 0$ $\newline$ $k = 3$ $\newline$ $p = \frac{1}{40}$ $\newline$ What is the focus of the parabola? $\newline$ Substitute $h = 0$ , $k = 3$ and $p = \frac{1}{40}$ . $\newline$ $(0, 3 + \frac{1}{40}) = (0, \frac{121}{40})$ $\newline$ Focus of the parabola: $(0, \frac{121}{40})$

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