Precalculus!! Please help!!!?

Question

Write an equation for the conic section.

1. Parabola with vertex at (0,0) and directrix y = 6

2. Circle with center (1,1) and radius 5

3. Hyperbola with center (0,0), foci at (-2,0) and (2,0), and vertices at (-1,0) and (1,0)

Answer 1

Vertex at (0, 0) and directrix y = 6

There fore the parabola equation is x²  = 4ay

Vertex at (0, 0) and directrix y = 6 then

The focus is (o , 6)

The distance between focus(x1, y1) and vertex(x2, y2) are √[(x2 - x1)² +(y2 - y1)²]

Substitute x1 = 0, y1 = 0 , x2 =0 and y2 = 6

There fore the distance (a) is √[(0-0)² + (6-0)²]

a =√(0² + 6²)

  = √(0 + 6²)

  =√(6²)

 =6

There fore a = 6

Substitute a = 6 in the parabola x² = 4ay

x² = 4(6)y

There the parabola is x² = 24y.

Comments

The above  first solution of parabola with vertex (0,0) and directrix y = 6 is wrong.

see the correct solution, answered by david.

Answer 2

The center of the circle is (1,1)

The center of the radius is 5

The center (a,b)  and radius r then the equation of the circle : (x -a)^2 + (y - b)^2 = r^2

Substitute a = 1, b = 1 and r = 5

(x - 1)² +(y - 1)² = 5^2

x² - 2x + 1 + y² - 2y +1 = 25

x² + y² -2x -2y + 2 = 25

Subtract 25 from each side

x² + y² -2x -2y + 2 - 25 = 25 - 25

x² + y² -2x -2y -23 = 0

There fore  the equation of a circle : x² + y² -2x -2y - 23 = 0.

Answer 3

The center of the hyperbola : (0,0)

The foci of the hyperbola : (-2,0) and (2,0)

The vertices of the hyperbola : (-1, 0) and (1,0)

The center (0,0) then equation of the hyperbola : x² / a² - y² / b² =1

The foci of the hyperbola : (-c,0) and (c,0)

The vertices of the hyperbola : (-a,0) and (a,0)

But c² = a² + b²

From the given problem c = 2 and a = 1

Substitute c = 2 and a = 1 in the c² = a² + b²

2² = 1² + b²

4 = 1 + b²

Subtract 1 from each side

4 - 1 = 1 +b² - 1

3 = b² + 0

 b² = 3

Take square root to each side

b = -√(3) , √(3).

Substitute a² = 1, b² = 3 in the equation x² / a² - y² / b² = 1

There fore the equation of the hyperbola : x² / 1 - y² / 3 = 1.

Answer 4

1) The parabola vertex (0,0) and directrix y  = 6

If a parabola has vetical axis,the standard form of the equation parabola  is

Where p not equals to 0.(h ,k ) is vertex .Focus is (h , k +p ) and directrix is y  = k - p .

Vertex (0,0)

h  = 0 and k  = 0

Directrix y  = k - p  = 6

 0 - p  = 6

p  = -6

p  is negitive, the parabola opens downward.

Substitute (h ,k ) and 4p  values in standardform of parabola.

Equation of parabola is (x - 0)² = - 24(y - 0)

Parabola equation is x ²  = -24y .