The elements of coordinate geometry, by S. L. Loney, M.A.

334

COORDINATE GEOMETRY.

The equation to the axis of the parabola is therefore
x cos (90~ + 0) + y sin (90~ + )= K,
i.e. - x sin 6 + y cos 0 = K,. e. ax~ +P fy=K a2 +2 3 ag +/3f (6)
ez~. ax + 1PY = I,/a a+fi2 2J......... (6).
Again, the vertex is the point in which the xis (6)
meets the curve (2).
We have therefore to solve (6) and (2), i. e. (6) and
(ag+ f) + 2gx + 2fy + c = 0............. (7)
The solution of (6) and (7) therefore gives the required
coordinates of the vertex.
361. It was proved in Art. 224 that if PV be a
diameter of the parabola and Q V the ordinate to it drawn
through any point Q of the curve, so that Q V is parallel to
the tangent at P, and if 0 be the angle between the diameter
P V and the tangent at P, then
Q V2 = 4a cosec2. P V................ (1).
If QL be perpendicular to P V and QL' be perpendicular
to the tangent at P, we have
QL Q Vsin, and QL' P Vsin0,
so that (1) is QL2 = 4a cosec 0. QL'.
Hence the square of the perpendicular distance of any
point Q on the parabola from any diameter varies as the
perpendicular distance of Q from the tangent at the end of
the diameter.
Hence, if Ax + By + C 0 be the equation of any
diameter and A'x + B'y + C'= 0 be the equation of the
tangent at its end, the equation to the parabola is
(Ax + By + C)2= (A'x + B'y C').........(2),
where X is some constant.
Conversely, if the equation to a parabola can be reduced
to the form (2), then
Ax + By+ C 0..................... (3)