Parabola | Problem 7 | DETERMINE THE VERTEX, FOCUS, DIRECTRIX, AND LATUS RECTUM | Judd Hernandez |

This video will focus on determining the vertex, focus, directrix, and latus rectum from a given parabola equation. The first step in solving such equations is to complete the square.

If the equation has a y-squared term, the parabola opens either to the left or to the right. If it has an x-squared term, it opens either upward or downward. The specific direction of opening (left/right or up/down) is determined by the sign of the constant when the equation is in standard form.

The video demonstrates completing the square for the equation \(y^2 - 5x + 12y = -16\). After rearrangement and completing the square for the y-terms, the equation becomes \((y+6)^2 = 5(x+4)\). This matches the standard form \((y-k)^2 = 4c(x-h)\).

From the standard form, the vertex (h, k) is identified as (-4, -6). Since the y-term is squared and the right side of the equation (5) is positive, the parabola opens to the right.

The term 4c in the standard form is equal to 5. Therefore, \(c = 5/4 = 1.25\). The length of the latus rectum (LR) is \(|4c|\), which in this case is \(5\) units.

Since the parabola opens to the right, the focus will be to the right of the vertex. The y-coordinate of the focus will be the same as the vertex (-6). The x-coordinate is found by adding 'c' to the x-coordinate of the vertex: \(-4 + 1.25 = -2.75\). So, the focus is at \((-2.75, -6)\) or \((-11/4, -6))\).

The directrix is a vertical line for parabolas opening horizontally, located 'c' units from the vertex in the opposite direction of the focus. So, \(x = -4 - 1.25 = -5.25\). The directrix is \(x = -5.25\) or \(x = -21/4)\).

By completing the square, the standard equation of the parabola was found to be \((y + 6)^2 = 5(x + 4))\. From this, the vertex is \((-4, -6))\), the focus is \((-2.75, -6))\), the directrix is \(x = -5.25))\), and the length of the latus rectum is \(5\) units.