Find an equation of the parabola with focus and directrix .

Question

Find an equation of the parabola with focus  and directrix  .

Quizplus · Accepted Answer

The vertex of the parabola is the midpoint between the focus and the directrix. Therefore, the vertex has coordinates $(0, \frac{1}{2}(0+4))=(0,2)$. Since the directrix is a horizontal line, the parabola opens upwards. Let $p$ be the distance between the vertex and the focus. This is also the distance between the vertex and the directrix. Thus, $p=2$. Using the definition of a parabola, we have $(x-0)^2+(y-2)^2=(2)^2+((y-4)^2$. Simplifying gives $x^2+(y-2)^2=4+(y-4)^2$, which expands to $x^2+y^2-4y+0x-8=0$. Completing the square gives $(x-0)^2+(y-2)^2=16$, which is the standard form of a parabola with vertex $(0,2)$ and $p=2$. Thus, the equation of the parabola is $x^2+(y-2)^2=16-y^2+4y$. Distributing the negative $y^2$ and combining like terms gives $x^2+2y^2-4y-16=0$. The only choice that matches this answer is $\boxed{	extbf{(B)}}$

Find an Equation of the Parabola with Focus and Directrix