Find the equations of the parabola whose directrix x = −4, axis y + 2 = 0, and latus rectum 10.

Question

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Answer 1

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Answer 2

Let focus F(b, - 2), then p = (b + 4)/2 and vertex of parabola at point ((b - 4)/2, - 2);

Eqyation of parabola (y + 2)² = 2(b + 4)(x - (b - 4)/2). If x = b then (y + 2)² = 2(b + 4)(b + 4)/2;

y + 2 = b + 4; y = b + 2; y + 2 = - (b + 4); y = - b - 6; Latus rectum : b + 2 - (- b - 6) = 10; 2b = 2; b = 1

Equation of parabola: (y + 2)² = 2(1 + 4)(x+3/2); (y + 2)² = 10(x + 3/2)

Answer 3

Because the directrix is vertical(x = -4) and the axis is horizontal (y=-2). The parabola will have the form:

(y-k)² = 4a(x-h)

The Latus Rectum = 4a = 10, so a = 5/2 (a is the focal length)

The axis (y=-2) indicates that the vertex point has a y value of -2

The vertex is 5/2 ( or a) to the right of the directrix (as is the focal point and the LR that goes through it are to the right of the vertex). The x value of the vertex is -4 + 5/2 = -3/2 So, (h,k) = (-3/2,-2)

(y+2)² = 10(x+3/2)

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Answer 4

directrix: x=-4

axis of symmetry: y=-2

latus rectum = 10

focus & vertex have y coordinates =-2

it's rightward opening parabola

vertically, directly above & below the focus are points on the parabola with y coordinates = -2+/-5= 3 and -7

use the point above (x,3). The distance from (x,3) to the directrix =5 = the distance from the focus

the x coordinate of that point on the parabola = 1, (1,3) is a point on the parabola,

focus= (1,-2) vertex =(-3/2,-2)

the parabola is y=a(x-h)^2 +k with (h,k)= vertex = (-3/2,-2)

x=a(y+2)^2 -3/2, plug in any point on the parabola to solve for a, such as (1,3)

1=a(3+2)^2 -3/2

5/2=25a

1/2=5a

a=1/10

x=(1/10)(y+2)^2 -3/2

check the answer

plug in the (1,3)

1=(1/10)(3+2)^2-3/2

1=25/10-3/2

1 =5/2-3/2=1

plug in (1,-7)

1=(1/10)(-7+2)^2-3/2

1=(1/10)(-5)^2-3/2

1=25/10-3/2

1=1