Question

when the focus and directrix are used to derive the equation of a parabola, two distances were set equal to each other.\\(\sqrt{(x - x)^2 + (y - (-p))^2} = \sqrt{(x - 0)^2 + (y - p)^2}\\)\
the distance between the directrix and a point on the parabola is set equal to the distance between the dropdown and the same point on the parabola.\
dropdown options: directix, focus, origin, x - axis\
(image shows a parabola with focus f(0, p), point p(x, y) on the parabola, directrix y = -p, and point q(x, -p))

Explanation:

To derive the equation of a parabola, we use the definition that any point \( P(x, y) \) on the parabola is equidistant from the focus and the directrix. The left - hand side of the equation \( \sqrt{(x - x)^{2}+(y-(-p))^{2}} \) represents the distance between the point \( P(x,y) \) on the parabola and the directrix (since the directrix is \( y=-p \), and the point \( Q(x, - p) \) is on the directrix, so the distance from \( P(x,y) \) to \( Q(x,-p) \) is calculated by the distance formula, and here \( (x - x)=0 \), so it simplifies to \( \sqrt{(y + p)^{2}}=\vert y + p\vert \)). The right - hand side \( \sqrt{(x - 0)^{2}+(y - p)^{2}} \) represents the distance between the point \( P(x,y) \) on the parabola and the focus \( F(0,p) \). So according to the definition of a parabola (the set of all points equidistant from a fixed point (focus) and a fixed line (directrix)), the distance between the directrix and a point on the parabola is set equal to the distance between the focus and the same point on the parabola.

Answer:

focus