Answer to Question #173398 in Calculus for Jon jay Mendoza

Question

Find the standard equation of the parabola which satisfies the given conditions.

1. Focus(-2, -5), directrix x=6
2. Vertex(-4, 2), focus(-4,-1)
3. Vertex(-8,3), directrix x= -10.5
4. Focus(7, 11), directrix y=4

Answer 1

$1. \ \ (x+2)² + (y+5)² = (y+6)²\\ x² + 4x + 4 + y² + 10y + 25 = y² + 12y + 36\\ x² + 4x -7 + y² - y² - 2y = 0\\ x² + 4x - 2y - 7 = 0$

2. The parabola has its vertex at (-4, 2) and focus at (-4, −1).

The parabola is of the type x² = −4by, with its vertex shifted from the origin to (-4, 2).

The distance from the vertex to the focus is b = 2 −(−1) = 3

⇒ The equation of the parabola is $(x+4)² = -4(3)(y−1).$

$(x−4)² = −12(y−1)$

3. The basic equations are:

$(x - h)² = 4p(y - k)$

and

$(y - k)² = 4p(x - h)$

When I sketch the vertex and the directrix, this shows that the parabola opens to the right. This means it would use this version of the equation.

(y - k)² = 4p (x - h)

The vertex is (h,k) The p value can be found by the distance between the vertex and the directrix, which is 2.5. The 4p value is negative because it opens to the left.klkkkk Plugging all this in.

(y - (-8))² = -4(1) (x - 2)

Simplifying

y² + 16x + 64 = -4(x - 2)

y² + 24x + 72 = 0

$4.\ \ \ \ (x-7)² + (y- 11)² = (y+4)²\\ x² -14x + 4 9+ y² -22y + 121= y² + 8y+ 16\\ x² - 1 4x+154 - 30y = 0\\$