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

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(B) The equation of a normal to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx - 2am - am^3$. Substituting $a=1$ and the point $(3, 0)$,we get $0 =

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obtuse angled triangle

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(B) The equation of a normal to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx - 2am - am^3$. Substituting $a=1$ and the point $(3, 0)$,we get $0 = 3m - 2m - m^3$. $m^3 - m = 0 \Rightarrow m(m-1)(m+1) = 0$. Thus,the slopes of the normals are $m = 0, 1, -1$. The coordinates of the feet of the normals are given by $(am^2, -2am)$. For $m=0$,$P = (0, 0)$. For $m=1$,$Q = (1, -2)$. For $m=-1$,$R = (1, 2)$. Calculating 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 $P, Q, R$ are drawn to the parabola $y^2 = 4x$ which intersect at the point $(3, 0)$. Then,$\Delta PQR$ is:

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