Answer to Question #171139 in Analytic Geometry 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. The distance of the directrix x=6 to the focus is 8 units therefore, 2p=-8, p=-4. Thus, the vertex is (2,-5).

The axis of symmetry is parallel to the x-axis:

(y - k)² = 4p(x - h) h=2 and k=-5

(y-(-5))²= 4(-4)(x-2)

(y+5)² = -16(x-2)

2. $(x-h)^2=4p(y-k)$

h=-4

k=2

p=-1-2=-3

$(x+4)^2=-12(y-2)$

$x^2+8x+16=-12y+24$

$y=-x^2/12-2/3x+2/3$

3. (y - k)²= 4p(x - h) h=-8 and k=3

k-p=-10.5

p=10.5-8=2.5

$(y-3)^2=10(x+8)$

4.$\sqrt{(y-k)^2 }=\sqrt{(x-a)^2+(y-b)^2}$

k=4

a=7

b=11

$\sqrt{(y-4)^2 }=\sqrt{(x-7)^2+(y-11)^2}$

$(y-4)^2 =(x-7)^2+(y-11)^2$

$y^2-8y+16=x^2-14x+49+y^2-22y+121$

$14y=x^2-14x+154$

$y=x^2/14-x+11$