I attend high school in Australia, but i often struggle with algebra (coz i keep making stupid mistakes all the time, why though? is it really lack of practice?)

(btw, do i need to register or anything?

Anyways, I've got a problem on parabola. We started locus 2 lessons ago, and a friend of mine came up with the idea of tilted parabolas (well, basicly a parabola with its axes rotated)

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and I tried to derive a formula for parabolas with vertex at the origin, and directrix at ax+by+c=0

I figured that, the coordinates of the focus of the parabola should be

x=(a*c)/(a^2+b^2), y=(b*c)/(a^2+b^2)

as
the focus lies on the normal of the directrix

and

the distance from the vertex to focus = distance to the directrix

then i did some algebra work (and i guess i didn't make any mistakes during the process

the equ for such an parabola came up to be

4c(ax+by)+2abxy-a^2*y^2-b^2*x^2
---------------------------------------------------------- = 0
a^2+b^2

when i fiddled with this formula,
i tried to place the directrix at

0=0x-y-1

which is equivalent to y=-1
(i.e the equ of the parabola should be x^2=4y

the formula seemed to work, however,
when i located the directrix at

x+y+1=0

it no longer forms a function, and i do not know how to graph non-functions on computer with Mupad, so i tried to use QBASIC to make a graph, however, the resultant graph seemed ... weird, it looked like 2 ordinary concave up parabolas. It could have been the way i used to obtain the y value, or is it that the formula is wrong?

well, by any chance, if anyone has the time, please tell me what is wrong with the formula. or the program

DECLARE SUB getval (x!)
SCREEN 12
DEFDBL A-Z
bx = 320
by = 240

TYPE value
a AS DOUBLE
b AS DOUBLE
END TYPE

DIM SHARED aval AS value

FOR x = 0 TO 24 STEP .1
getval (x)
IF NOT (aval.a = 0 AND aval.b = 0) THEN
PSET (bx + x * 5, by + aval.a * 5)
PSET (bx + x * 5, by + aval.b * 5)
END IF
NEXT x

getval (0)
PRINT aval.c

DEFDBL A-Z
SUB getval (x)
'
' formula used
' b+-sqrt(b^2-4ac)
'
' evaluating b and c
'
' c = x^2-4x

qC = x ^ 2 - 4 * x

' b = -(4+2x)

qB = -(4 + 2 * x)

'
' evaluating discriminant
' a = 1

IF qB = 0 THEN
delta = 0 + 4 * qC
ELSE
delta = qB ^ 2 - 4 * qC
END IF

IF SGN(delta) = -1 THEN
aval.a = 0
aval.b = 0
EXIT SUB
END IF

aval.a = (-qC + SQR(delta)) / 2
aval.b = (-qC - SQR(delta)) / 2

END SUB

(btw, I actually fouad this site while searching for materials on the subject "Beauty of Math", i am doing a speech tomorrow on that topic, so better get going./ (Well, I am just trying to convince my english teacher and also myself, that, Math is beautiful.

thank you ppl

No, you do not.

>x=(a*c)/(a^2 b^2), y=(b*c)/(a^2 b^2)

This is correct.

>4c(ax by) 2abxy-a^2*y^2-b^2*x^2
>---------------------------------------------------------- =
>0
> a^2 b^2

This is correct, except you may remove the positive denominator.

>SUB getval (x)
> '
> ' formula used
> ' b -sqrt(b^2-4ac)
> '

But what are a, b, c, and x?

Assume you try solving Ay² - By + C = 0. Then the formula will be

y_1,2 = (B ± sqrt(B² - 4AC))/2A.

In your case,

A = a²,
B = 4bc + 2abx,
C = b²x² - 4acx,

where a, b, c, and x are as they are used in your problem Description.