maths
posted by lia .
The parabola y2 = 4ax, where a > 0, and the rectangular hyperbola xy = C2, where
C > 0, intersect at right angles. Show that the tangent and normal to either curve at the
point of intersection meet the xaxis at T and N where TN = 2pa, where p is an integer
to be determined.

from y^2 = 4ax
2y dy/dx = 4a
dy/dx = 2a/y  >slope of tangent to parabola
from xy = c^2
xdy/dx + y = 0
dy/dx = y/x > slope of tangent of hyperbola
but we are told that they intersect at right angles, so the tangents must be perpendicular, making
2a/y = x/y
x = 2a
sub into xy=c^2
y = c^2/(2a)
so the intersect at (2a, c^2/(2a)
(Now it gets messy)
at parabola, slope = 2a/(c^2/2a) = 4a^2/c^2
equation of tangent:
y  c^2/(2a) = (4a^2/c^2)(x2a)
yc^2  c^4/(2a) = 4a^2x  8a^3
at the xaxis, y = 0
 c^4/(2a) = 4a^2x  8a^3
4a^2x = 8a^3  c^4/(2a)
x = 2a  c^4/(8a^3) > the xintercept, call it T
at the hyperbola, slope =
slope = y/x = (c^2/(2a))/(2a) = c^4/(4a^2)
notice that this is the negative reciprocal of the slope at the parabola, so far so good!
equation of tangent:
(y  c^2/(2a)) = (c^2/(4a^2)) (x  2a)
4a^2y  2ac^2 = c^2x + 2ac^2
again at the xintercept , let y = 0
 2ac^2 = c^2x + 2ac^2
c^2 x = 4ac^2
x = 4a > the other xintercept, call it N
so TN = 4a  (2a  c^4/(8a^3)) = 2a + c^4/(8a^3)
but TN = 2pa
so
2a + c^4/(8a^3) = 2pa
p =1 + c^4/(16a^4)
????????? What do you think????
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