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Problem
Problem
Prove that the area of the triangle inscribed in the parabola
y
2
=
4
a
x
y^2 = 4ax
y
2
=
4
a
x
is
1
8
a
(
y
1
−
y
2
)
(
y
2
−
y
3
)
(
y
3
−
y
1
)
,
where
y
1
,
y
2
,
y
3
are the ordinates of the vertices.
\frac{1}{8a} (y_1 - y_2)(y_2 - y_3)(y_3 - y_1), \text{ where } y_1, y_2, y_3 \text{ are the ordinates of the vertices.}
8
a
1
(
y
1
−
y
2
)
(
y
2
−
y
3
)
(
y
3
−
y
1
)
,
where
y
1
,
y
2
,
y
3
are the ordinates of the vertices.
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