derive the equation of the parabolla with its vertex on the line 7x+3y-4=0 and containing ponts (3,-5) and (3/2,1) the axis being horizontal.

Answer 1

Not possible
If the vertex is on the line 7x + 3y - 4 = 0, which has a slope of -7/3,
then its axis would have to have a slope of 3/7, which is NOT horizontal.
Check your question.

Answer 2

the equation is

x = a(y-k)^2 + h

The two points given mean that

3-h = a(-5-k)^2
3/2 - h = a(1-k)^2

Divide to get rid of the a, and we have

(3-h)/(5+k)^2 = (3/2 - h)/(1-k)^2
or
2(17+3k)(1-k)^2 = (13+6k)(5+k)^2
and
k=1
so, h=1

x-1 = a(y+1)
3-1 = a(-5+1)
a = -1/2

finally,

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

Answer 3

Misunderstood the question
Go with Steve's answer.

Answer 4

actually, there is a second solution, where k = -97/17, h = 359/119, making a = -1/42

and

x = -1/42 (y + 97/17) + 359/119

(I think)

Answer 5

actually, I botched it near the end:

(h,k) = (1,-1) as solved, but

x-1 = a(y+1)^2
3-1 = a(-4)^2
2 = 16a
a = 1/8

x = 1/8(y+1)^2 + 1

Answer 6

also, the second solution is correctly expressed as

x = -17/504 * (y+97/17)^2 + 359/119

plotting the three graphs shows that the two vertices lie on the line, go through both given points, and open in opposite directions.

*whew*