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October 31, 2014

October 31, 2014

Posted by **lia** on Wednesday, March 23, 2011 at 7:04am.

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.

- maths -
**Reiny**, Wednesday, March 23, 2011 at 1:09pmfrom 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????

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