19ans =1
(1) The sum of X-coordinate of the point of Intersection of the curves
x^{2}=x+y+4 and y^{2}=y-15x+36 in X-Y plane.
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4 Answers
1eliminate y from (1) & substitute in (2)
u vl get an eqn in 4th degree....
now sum of roots is = -(coeff of x3)/(coeff of x4)
106from (i) y = x2-x-4
putting this in (ii)
x4 + x2 + 16 - 2x3 + 8x - 8x2 = x2-x-4 - 15x + 36
=> x4 - 2x3 - 8x2 + 24x - 16 = 0
we need to find the sum of REAL roots (there maybe imaginary roots as well)
the above exp is factorised as
(x-2)2(x2+2x-4) = 0 (both sides give real roots)
So the sum of roots is 2 (from (x-2)2) + (-2) (from(x2+2x-4)
so the sum of roots is zero
so the sum of x-coordinates of pts of interseciton is zero.