Answer to Question #157470 in Analytic Geometry for Ariel

Question

Write down the standard form of the equation of a parabola in a given directrix and vertex at (0, 0)
(a) x = -4
(b) y = 2
(c) y = - 3

Answer 1

We can write the equation of parabola in the form:

$4p(y-k)=(x-h)^2$ . The focus is (h,k+p) and the directrix is $y=k-p$. The vertex is (h,k). In item b) we receive: $k-p=2$ and $h=0, k=0$. Thus, $p=-2$. The equation is: $-8y=x^2$

In c). we get $k-p=-3$, $h=0,k=0$. The equation is: $12y=x^2$

Now we write the equation in the form: $4p(x-k)=(y-h)^2$ . The focus is (h,k+p) and the directrix is $x=k-p$. The vertex is (h,k). For a). we receive: $k=0,h=0$, $k-p=-4$. Thus, $p=4.$ The equation is: $16x=y^2$ . The standard forms of equations of parabola are: $y=ax^2+bx+c$ and $x=ey^2+fy+g$ , where $a,b,c,e,f,g\in{\mathbb{R}}$. All obtained equations are in standard forms.

Answer: a). $16x=y^2$; b). $-8y=x^2$ ; c). $12y=x^2$