parabolas 1

Question

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.

Answer 1

Let ( x₀ , y₀ ) be any point on the parabola. Find the distance between (x₀ , y₀) and the focus. Then find the distance between (x₀ , y₀) and directrix.

Equate these two distance equations and the simplified equation in x₀ and y₀ is equation of the parabola.

The distance between (x₀ , y₀) and (0, 8) is √(0-x₀)^2+(8-y₀ )^2

The distance between (x₀ , y₀) and the directrix, y =- 8 is | y₀+ 8|.

Equate the two distance expressions and square on both sides.

√(0-x₀)^2+(8-y₀ )^2 =| y₀+ 8|.

Square on both sides

(0-x₀)^2+(8-y₀ )^2 =| y₀+ 8|^2

x₀^2+64+y₀^2-16y₀= y₀^2+16y₀+64

Simplify :

x₀^2= 32y₀

This equation in (x₀ , y₀) is true for all other values on the parabola and hence we can rewrite with (x , y).

So, the equation of the parabola with focus (0, 8) and directrix is y = -8 is x^2= 32y

 .