Question

Difficulty: HardConic Sections

A parabola in the standard (x,y)(x, y) coordinate plane is defined by the equation x26x8y+25=0x^2 - 6x - 8y + 25 = 0. What is the distance, in coordinate units, between the focus and the directrix of this parabola?

Answer: 4

Answer

The distance between the focus and the directrix of the parabola is 4.
By completing the square on the equation x26x8y+25=0x^2 - 6x - 8y + 25 = 0, we get (x3)2=8(y2)(x-3)^2 = 8(y-2). Since the coefficient of the linear factor is 88, we set 4p=84p = 8, which yields p=2p = 2. The distance from the focus to the directrix is 2p=2(2)=42p = 2(2) = 4.

Step-by-Step Solution

1
Isolate the terms containing xx on one side of the equation.
x26x=8y25x^2 - 6x = 8y - 25
To set up the equation for completing the square on the xx terms.
2
Complete the square for the quadratic expression in xx by adding 99 to both sides.
x26x+9=8y16    (x3)2=8y16x^2 - 6x + 9 = 8y - 16 \implies (x-3)^2 = 8y - 16
Adding (6/2)2=9( -6/2 )^2 = 9 creates a perfect square trinomial on the left side.
3
Factor out the coefficient of yy on the right side to write the equation in standard form.
(x3)2=8(y2)(x-3)^2 = 8(y-2)
This matches the standard form equation (xh)2=4p(yk)(x-h)^2 = 4p(y-k) for a vertical parabola.
4
Determine the value of the focal parameter pp from the standard form.
4p=8    p=24p = 8 \implies p = 2
Comparing the standard form coefficient 4p4p with the value 88 gives p=2p = 2.
5
Calculate the total distance between the focus and the directrix.
2p=2(2)=42p = 2(2) = 4
The vertex is situated halfway between the focus and the directrix, making the distance between them 2p2p.

Key Concept

Finding the geometric properties of a parabola by completing the square to convert its general equation to standard form.
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