Question

The focus of a parabola is ($1,5$) and its directrix is $x+y+2=0$. Find the equation of the parabola. Its vertex and length of latus rectum ?

Answer 1

The definition of a parabola is the set of points (x,y) such that the distance from the point to the directrix is equal to the distance from the point to the focus
The distance from the point (x,y) to the point (1,5) is sqrt((x -1)^2 + (y - 5)^2)
The distance from point (x,y) to the line x + y + 2 = 0 is |x + y + 2|/sqrt2
(x + y + 2)^2/2 = (x - 1)^2 + (y - 5)^2
x^2 + y^2 + 4 + 4x + 4y + 2xy = 2(x^2 - 2x + 1 + y^2 - 10y + 25)
0 = x^2 + y^2 - 2xy - 8x - 24y + 48, equation of parabola
The vertex is the midpoint of the line along the axis of symmetry from the focus to the directrix.
Eqn of directrix y = -x - 2, slope -1
Slope of axis is 1, passes through (1,5) so its eqn is y - 5 = x - 1
These intersect where (-x - 2) - 5 = x - 1
-x - 7 = x - 1
2x = -6
x = -3
y = 1
The vertex is the midpoint of the line between (-3,1) and (1,5)
[(-3 + 1)/2 , (1 + 5)/2] = (-1,3) vertex of parabola
The length of the latus rectum is found from the distance between its endpoints,
The latus rectum is parallel to the directrix and passes through (1,5)
y - 5 = -1(x - 1)
y = 6 - x
Substitute into the eqn of the parabola for y and solve for x, two points result as the endpoints of the latus rectum
(-3,9) and (5,1)
The length of the latus rectum is
sqrt((-3 - 5)^2 + (9 - 1)^2)
= sqrt(8^2 + 8^2)
= 8sqrt2