The parabola y2 = 4ax, where a > 0, and the rectangular hyperbola xy = C2, where

Question

C > 0, intersect at right angles. Show that the tangent and normal to either curve at the
point of intersection meet the x-axis at T and N where TN = 2pa, where p is an integer
to be determined.

Answer 1

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)(x-2a)
yc^2 - c^4/(2a) = 4a^2x - 8a^3
at the x-axis, y = 0
- c^4/(2a) = 4a^2x - 8a^3
4a^2x = 8a^3 - c^4/(2a)
x = 2a - c^4/(8a^3) -----> the x-intercept, 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 x-intercept , 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????