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Analytical Geometry

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Joined:	15 Feb 2007
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16 Jul 2007 17:34:38 IST

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challenging problem 4

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if the nrmal to the curve y=x^2 at the point P,Q and R pass thruogh the point (0,3/2) , find the radius of the circle circumscribing the triangle PQR.

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Titun

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Joined: 23 Dec 2006

Posts: 374

16 Jul 2007 18:33:12 IST

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y = x²

Therefore, Slope of normal = - dx / dy = - 1 / 2x

Therefore, equation of normal at any point (x,y) lying on the parabola and passing through (0, 3/2) is,

y - 3 / 2 = - 1 / 2x . (x - 0)

i.e x² - 3 / 2 = - 1 / 2x . x

i.e 2x³ - 2x = 0 i.e x³ - x = 0 i.e x ( x - 1 ) (x + 1) = 0

So, x = 0 , 1 , - 1

Consequently, y = 0 , 1 , 1

So, coordinates of P = (0,0)
coordinates of Q = (-1,1)
coordinates of R = (1,1)

Since, slope of PQ x slope of PR = 1x (-1) = -1

Therefore, PQR is a right angle triangle and angle QPR is a right angle.
The circumcentre of right angled triangle PQR is at the midpoint of the hypotenuse QR i.e (0, 1)

So, the circumradius is the distance between any of the vertices and the circumcentre = 1 unit.

Ans : 1 unit

Cheers !!!!!

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