Question

Prove that the equation to the parabola, whose vertex and focus are on the axis of x at distances a and a' from the origin respectively, is $y^2 = 4(a' - a)(x - a)$.

Answer 1

This is a proof question, so no specific answer choice is applicable. The derived equation is $y^2 = 4(a'-a)(x-a)$.

Answer 2

Let the vertex of the parabola be $V$ and the focus be $F$. Given that $V$ and $F$ are on the x-axis at distances $a$ and $a'$ from the origin, their coordinates are $V(a, 0)$ and $F(a', 0)$. Since both vertex and focus lie on the x-axis, the axis of the parabola is the x-axis. The standard equation of a parabola with vertex at $(h, k)$ and axis parallel to the x-axis is $(y-k)^2 = 4c(x-h)$, where $c$ is the focal length (distance from vertex to focus).

In this case, the vertex is $(h, k) = (a, 0)$. The focal length $c$ is the distance between the vertex $V(a, 0)$ and the focus $F(a', 0)$, which is $c = |a' - a|$. For the parabola to open towards the positive x-axis (as implied by the form $x-a$), we take $c = a' - a$. If $a' < a$, the parabola would open towards the negative x-axis, and the equation would be $(y-0)^2 = 4(a-a')(x-a')$, with vertex at $(a',0)$ and focus at $(a,0)$. Assuming $a' > a$, the focal length is $a'-a$.

Substituting $h=a$, $k=0$, and $c=a'-a$ into the standard equation: $(y-0)^2 = 4(a'-a)(x-a)$ $y^2 = 4(a'-a)(x-a)$

This proves the given equation.