cartesian form of the parabola?

Question

find the equation of the parabola with its axis of symetry parrallel to the x-axis and which passes throught the points (3,3), (6,5) and (6,-3)

help?

Answer 1

Knowing two points on a vertical line is very helpful.

(6,5) and (6,-3) must be equidistant from the line of symmetry, so average the y coordinates:

[ 5 + (-3) ] / 2 = 2 / 2 = 1

The line of symmetry is y=1.

Now we know that it is of the form:

x = a(y-1)² + b

Plug in (6,5) and (3,3)

6 = a(5-1)² + b
3 = a(3-1)² + b

16a + b = 6
4a + b = 3

subtract the two equations:

12a = 3
a = 3/12 = 1/4

b = 6 - 16a = 6 - 16/4 = 6 - 4 =2

So the equation is:

x = (y-1)²/4 + 2

or you can expand:

x = y²/4 - y/2 + 9/4 = (y² - 2y + 9)/4

How you write it is up to you (or your teacher, anyway).

Answer 2

The equation of a horizontal parabola is:

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

h = x-coordinate of vertex

k = y-coordinate of vertex

So anyway... the first thing you should realize is that the points on the same x-coordinate have the exact same distance from the axis of symmetry.

So when you look at the 2 points, (6,5) and (6,-3)

You know that the axis of symmetry must be y = ?, where ? = midpoint of the two Y-COORDINATES

So the axis of symmetry is:

y = (5-3)/2 = 1

The vertex must only have 1 possible y-coordinate, which is 1.

4p(x-h) = (y-1)^2

Plug in x and y.

4p(3-h) = 4

Divide by 4...

p(3-h) = 1

Plug in another 1.

4p(6-h) = 16

Divide by 4...

p(6-h) = 4

We do have enough info now.

Factor everything though lol.

3p - ph = 1

6p - ph = 4

Subtract both sides.

-3p = -3

p = 1

So p is 1.

Without a doubt h is 2.

So the vertex is (2,1)

Foci = 1

_______________________________

Equation.....

4(x-2) = (y-1)^2

Answer 3

Find the equation of the parabola with its axis of symmetry parallel to the x-axis and which passes throught the points (3,3), (6,5) and (6,-3).

This is a sideways parabola with equation of the form:

ay² + by + c = x

We have three equations and three unknowns.
(3,3), (6,5) and (6,-3).

9a + 3b + c = 3
25a + 5b + c = 6
9a - 3b + c = 6

Solving we get

a = 1/4
b = -1/2
c = 9/4

So the equation of the parabola is:

x = y²/4 - y/2 + 9/4

Answer 4

equation of parabola whoose axis is parallel to x axis is
x= ay^2 +by +c-------------------- equation a
now substitute the points in the above eqation

(3,3) 3 = a 3^2 +3b +c

3= 9a +3b +c --------------------------------equation 1

(6,5) 6= a 5^2 +5b +c

6 = 25a +5b +c --------------------------------equation 2

(6,-3) 6 = a 3^2 -3b+c

6 = 9a -3b +c --------------------------------equation 3

solve equation 1 2 and 3 we get the values of a b c
a = 1/4
b = -1/2
c = 9/4

So the equation of the parabola becomes:

x = y²/4 - y/2 + 9/4