The locus of the orthocentre of the triangle | vedclass

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Question

(A) Let the focal chord be $PQ$ where $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$. Since $PQ$ is a focal chord,$t_1t_2 = -1$.The normals at $P$ and $Q$

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$y^2 = a(x - 3a)$

Vedclass · Answer

(A) Let the focal chord be $PQ$ where $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$. Since $PQ$ is a focal chord,$t_1t_2 = -1$.The normals at $P$ and $Q$ intersect at $R(h, k)$,where $h = a(t_1^2 + t_2^2 + t_1t_2 + 2)$ and $k = -at_1t_2(t_1 + t_2)$.Substituting $t_1t_2 = -1$ into the expressions:$h = a(t_1^2 + t_2^2 - 1 + 2) = a(t_1^2 + t_2^2 + 1)$$k = -a(-1)(t_1 + t_2) = a(t_1 + t_2)$Now,$(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2 = t_1^2 + t_2^2 - 2$.So,$t_1^2 + t_2^2 = (t_1 + t_2)^2 + 2 = (k/a)^2 + 2$.Substituting this into the expression for $h$:$h = a((k/a)^2 + 2 + 1) = a(k^2/a^2 + 3) = k^2/a + 3a$.Rearranging for $k^2$:$k^2 = a(h - 3a)$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = a(x - 3a)$.

The locus of the orthocentre of the triangle formed by the focal chord of the parabola $y^2 = 4ax$ and the normals drawn at its extremities is

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