Respuesta :

calculista

Answer:

[tex]y=\frac{1}{14}x^2[/tex]

Step-by-step explanation:

we know that

The directrix of the parabola is perpendicular to the axis of symmetry of the parabola

In this problem  the directrix is y=-3.5

so

The axis of symmetry is parallel to the y-axis

we have a vertical parabola

Also, the vertex is at the origin

That means-----> the parabola open upward

The equation of a vertical parabola can be written as

[tex]x^2=4py[/tex]

where

p is the distance between the vertex and the directrix

In this problem

the distance between the vertex and the directrix is 3.5

[tex]p=3.5[/tex] ----> the value of p is positive because the parabola open upward

substitute

[tex]x^2=4(3.5)y[/tex]

[tex]x^2=14y[/tex]

isolate the variable y

[tex]y=\frac{1}{14}x^2[/tex]

GeekySpitz

For the Quadratic Relations and Conic Sections Unit Test Part 1:

1. C. Ellipse, Domain: {-1≤x≤1}, Range: {-2≤y≤2}

2. D. y=x^2/5

3. B. y=1/14 x^2

4. B. 5 inches

5. A. x=1/12 (y-4)^2+4

6. A. (x-3)^2+(y-4)^2=36

7. B. (x+8)^2+(y+4)^2=25

8. A. (x+3)^2+(y+6)^2=16

9. A. Center at (7,-6): radius 4

10. B. x^2/25+y^2/169=1

11. A. (0, ±6)

12. C. 10 feet

13. C. Conic section is on the x axis, should be two on the x axis, A and C, NOT the rounded one.

14. D. (0, ±5)

REMEMBER check your test to see if answers match up!!!

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