What is the latus rectum for 6y squared plus 24y plus 25?

Answer 1

The question is incomplete because it is not an equation.

Assume that the expression given equals x, i.e.:

6*y^2+24*y+25=x.

Completing the square on y:

6*[y^2 + 4*y + (+4/2)^2 - (+4/2)^2] + 25 = x,

6*[(y+2)^2 - 4] + 25 = x,

6*(y+2)^2 - 24 + 25 = x,

6*(y+2)^2 = x-1, and

(y+2)^2 = (1/6)*(x-1).

This is a translated parabola with y=-2 its axis of symmetry and (1,-2) its vertex. The domain of the parabola is x >= 1 and the range is that y can be any real number.

If the distance from the vertex to the focus along the axis of symmetry is called p, then, from the equation, 4*p = 1/6 and p = 1/24. That puts the focus at (1+1/24,-2) and the directrix at x = 1-1/24.

The length of the line segment called the latus rectum is |4*p| = 1/6 with endpoints (1+1/24,-2+1/12) and (1+1/24,-2-1/12).

The vertex and the endpoints of the latus rectum are points on the parabola and, in conjunction with the domain and range, are used to sketch the parabola.

BTW, "latus" and "rectum" are latin for "side" and "to lead in a straight-line or in the right direction" (the human large intestine's last, and straight, section is also called the "rectum").