How do you write the equation given vertex (0,1) and focus (0,5)?

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Answer 1 · 2016-11-24T22:32:55

Please see the explanation for steps leading to the equation $y = \frac{1}{16} x^{2} + 1$ $y = 1/16x^2 + 1$

The y coordinate changes between the vertex and the focus, therefore, we know that the equation is of the form:

$y = f (x)$ $y = f(x)$

Note: If it were the x coordinate that changes between the vertex and the focus then the equation would be of the form $x = f (y)$ $x = f(y)$

The vertex form for this type is:

$y = a {(x - h)}^{2} + k$ $y = a(x - h)^2 + k$

where (h, k) is the focus and $a = \frac{1}{4 f}$ $a = 1/(4f)$ where f is the vertical shift from the vertex to the focus.

Substitute 0 for h and 1 for k:

$y = a {(x - 0)}^{2} + 1$ $y = a(x - 0)^2 + 1$

$y = a x^{2} + 1$ $y = ax^2 + 1$

The vertical shift from the vertex to the focus is:

$f = (5 - 1) = 4$ $f = (5 - 1) = 4$

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

$a = \frac{1}{16}$ $a = 1/16$

Substitute $\frac{1}{16}$ $1/16$ for a:

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