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Problem
Problem
If the three normals from a point to the parabola
y
2
=
4
a
x
y^2 = 4ax
y
2
=
4
a
x
cut the axis in points whose distance from the vertex are in AP, show that the point lies on the curve
27
a
y
2
=
2
(
x
−
2
a
)
3
27ay^2 = 2(x - 2a)^3
27
a
y
2
=
2
(
x
−
2
a
)
3
.
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