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August 30, 2014

August 30, 2014

Posted by **chris** on Monday, November 1, 2010 at 5:07am.

Tangents are drawn to a parabola x^2=4y

from an external point A(x1,y1) touching the parabola at P and Q

(a) Prove that the mid point, M, is the point (x1,0.5 x1^2-y1)

and

(b) if A moves along the straight line y=x-1 find the equation of the locus M

- math -
**Reiny**, Monday, November 1, 2010 at 9:05amFor ease of typing I am going to let

A(x1, y1) = A(a,b)

Let P(x,y) be the point of contact

slope of AP = (b-y)/(a-x)

but slope of tangent by Calculus ,(differentiate x^2 = 4y )

= x/2

so (b-y)/(a-x) = x/2

2b - 2y = ax - x^2

2b - 2(x^2/4) = ax - x^2

x^2 - 2ax + 4b = 0

by quadratic equation

x = (2a ± √(4a^2 - 16b)/2

= (a ± √(a^2 - 4b)

so P is ( (a + √(a^2 - 4b) , [(a + √(a^2 - 4b)]^2/4 )

and Q is ( (a - √(a^2 - 4b) , [(a - √(a^2 - 4b)]/4)

now remember how to find the midpoint?

for the x coordinate, add the 2 x's of the endpoints and divide by 2

clearly that would give us a

for the y coordinate, I will not type it all out, but when you expand the above y values of the endpoints and then add them up , the middle terms drop out and I got

y = (2a^2 - 4b)/4

= (a^2/2 - b)

M is (a, a^2/2 - b) or in your original version

M is (x1, x1^2/2 - y1) which is what we were supposed to prove.

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