arrebolyt

2022-01-28

What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y=-15?

Expert

Explanation:
Focus is at (14,5) and directrix is y=-15. Vertex is at midway
between focus and directrix. Therefore vertex is at
(14, $\frac{5-15}{2}$) or (14, -5). The vertex form of equation of
parabola is $y=a{\left(x-h\right)}^{2}+k$; (h.k); being vertex. Here
h=14 and k=-5 So the equation of parabola is
$y=a{\left(x-14\right)}^{2}-5$. Distance of vertex from directrix is
d=15-5=10, we know $d=\frac{1}{4|a|}\therefore |a|=\frac{1}{4d}$ or
$|a|=\frac{1}{4\cdot 10}=\frac{1}{40}$. Here the directrix is below
the vertex , so parabola opens upward and a is positive.
$\therefore a=\frac{1}{40}$ Hence the equation of parabola is
$y=\frac{1}{40}{\left(x-14\right)}^{2}-5$

Expert

Explanation:
the standard form of a parabola in translated form is.
${\left(x-h\right)}^{2}=4p\left(y-k\right)$
where (h,k) are the coordinates of the vertex
and p is the distance from the vertex to the focus
since the directrix is below the focus then the curve
opens upwards
coordinates of vertex $=\left(14,\frac{5-15}{2}\right)=\left(14,-5\right)$
and p=5-(-5)=10
$⇒{\left(x-14\right)}^{2}=40\left(y+5\right)←$ equation of parabola

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