Normals at P,Q,and R are drawn to the parabol | vedclass

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(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx - 2am - am^3$. Substituting $a=1$,we get $y = mx - 2m - m^3$.Since

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obtuse angled triangle

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(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx - 2am - am^3$. Substituting $a=1$,we get $y = mx - 2m - m^3$.Since the normal passes through $(3, 0)$,we have $0 = 3m - 2m - m^3$,which simplifies to $m^3 - m = 0$.Solving for $m$,we get $m(m-1)(m+1) = 0$,so $m = 0, 1, -1$.The coordinates of the points are given by $(am^2, -2am)$.For $m=0$,$P = (0, 0)$.For $m=1$,$Q = (1, -2)$.For $m=-1$,$R = (1, 2)$.Now,calculate the side lengths:$PQ = \sqrt{(1-0)^2 + (-2-0)^2} = \sqrt{1+4} = \sqrt{5}$.$PR = \sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1+4} = \sqrt{5}$.$QR = \sqrt{(1-1)^2 + (2-(-2))^2} = \sqrt{0+16} = 4$.Since $(QR)^2 = 16$ and $(PQ)^2 + (PR)^2 = 5 + 5 = 10$,we have $(QR)^2 > (PQ)^2 + (PR)^2$.Therefore,$\Delta PQR$ is an obtuse-angled triangle.

Normals at $P$,$Q$,and $R$ are drawn to the parabola $y^2 = 4x$ which intersect at the point $(3, 0)$. Then,the triangle $\Delta PQR$ is:

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