Question

find the differential equation of all parabolas vertices at the origin.

Answer 1

This one sounds tedious, especially since the focus is along the line y = x instead of the x or y axis!
For Step 1 below, drawing a picture is imperative.
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Step 1: Find the equation of such a parabola (lengthy!!)

Let the focus of one of these parabolas be at (x, y) = (a, a) for some a in [0, 5].

Since the vertex of a parabola must be at the origin, the directrix must be the
line with equation y = -x - 2a.
(Why? One, the vertex must be equidistant from the focus and the directrix.
So, the point (-a, -a) must be on the directrix. Secondly, it must be perpendicular
to the line y = x by this last remark. This yields the line quoted above.)

Now, the 'fun' part: Deducing the equation of this kind of parabola.
Let (xo, yo) be on the parabola.

Since the axis is y = x (with slope 1), a line passing through (xo, yo) is
y = x + (yo - xo). Now, we find the point of intersection with the directrix y = -x - 2a.
==> The point of intersection is ((xo - yo)/2 - a, (yo - xo)/2 - a).

With this line, we can now compute the distance D between (xo, yo) and the directrix,
being the length of the perpendicular between the point and the line:
==> D = sqrt[(xo - ((xo - yo)/2 - a))^2 + (yo - ((yo - xo)/2 - a))^2]
==> D = sqrt[((xo + yo)/2 + a)^2 + (yo + xo)/2 + a)^2]
==> D = sqrt [2((xo + yo)/2 + a)^2].

Next, we compute the distance D' between (xo, yo) and the focus (a, a).
This is much easier: D' = sqrt [(xo - a)^2 + (yo - a)^2].

Finally, the equation of the parabola is defined by the set of all points (xo, yo)
which satisfy D = D':
sqrt [2((xo + yo)/2 + a)^2] = sqrt [(xo - a)^2 + (yo - a)^2]

Dropping subscripts and simplifying, we get
2((x + y)/2 + a)^2 = (x - a)^2 + (y - a)^2
==> (2/4) * ((x + y) + 2a)^2 = (x - a)^2 + (y - a)^2
==> ((x + y) + 2a)^2 = 2 [(x - a)^2 + (y - a)^2]
==> (x^2 + 2xy + y^2) + 4a(x + y) + 4a^2 = 2 (x^2 - 2ax + y^2 - 2ay + 2a^2)
==> x^2 + y^2 - 2xy - 8ax - 8ay = 0.
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Step 2: Use Implicit Differentiation to find the slope of the tangent to such a parabola

Differentiating yields
2x + 2yy' - (2y + 2xy') - 8a - 8ay' = 0
==> (y - x - 4a)y' + (x - y - 4a) = 0.

Since a = (x^2 + y^2 - 2xy) / (8x + 8y), the equation becomes
(y - x - [(x^2 + y^2 - 2xy) / (2x + 2y)]) y' + (x - y - [(x^2 + y^2 - 2xy) / (2x + 2y)]) = 0
==> (2y^2 - 2x^2 - (x^2 + y^2 - 2xy)) y' + (2x^2 - 2y^2 - (x^2 + y^2 - 2xy)) = 0

Solving for y' yields
y' = -(x^2 - 3y^2 + 2xy) / (y^2 - 3x^2 + 2xy).

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Step 3: The Orthogonal family.

Curves orthogonal to these parabolas will have slopes which are negative reciprocals of the tangent line slopes from the parabola.

So, for the orthogonal curves:
y' = (y^2 - 3x^2 + 2xy)/(x^2 - 3y^2 + 2xy)

Now, we solve for y:
y' = [(y/x)^2 - 3 + 2(y/x)] / [1 - 3(y/x)^2 + 2(y/x)].

Let y = xv; dy/dx = v + x dv/dx.
==> v + x dv/dx = (v^2 - 3 + 2v) / (1 - 3v^2 + 2v)
==> x dv/dx = (3v^3 - v^2 + v - 3) / (1 - 3v^2 + 2v)

Separating variables:
(1 - 3v^2 + 2v) dv / (3v^3 - v^2 + v - 3) = dx/x
==> -(3v + 1) dv / (3v^2 + 2v + 3) = dx/x.

Integrating both sides:
(-1/2) ln(3v^2 + 2v + 3) = ln x + A
==> ln(3v^2 + 2v + 3) = -2 ln x + (-2A)
==> 3v^2 + 2v + 3 = C/x^2, where C = e^(-2A)

Letting v = y/x:
3(y/x)^2 + 2(y/x) + 3 = C/x^2
==> 3y^2 + 2xy + 3x^2 = C.

Final answer:
The orthogonal family of curves to the given parabolas is
3y^2 + 2xy + 3x^2 = C.

I hope this helps!