The tangent PT and the normal PN to the parab | vedclass

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Question

(C) Let the point $P$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$. It meets the axis $(y=0)$ at $T(-at^2, 0)$.The equation

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(C) Let the point $P$ be $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$. It meets the axis $(y=0)$ at $T(-at^2, 0)$.The equation of the normal at $P$ is $y = -tx + 2at + at^3$. It meets the axis $(y=0)$ at $N(2a + at^2, 0)$.Let the centroid of $	riangle PTN$ be $R(h, k)$.$h = \frac{at^2 - at^2 + 2a + at^2}{3} = \frac{2a + at^2}{3}$$k = \frac{2at + 0 + 0}{3} = \frac{2at}{3} \Rightarrow t = \frac{3k}{2a}$.Substituting $t$ in the expression for $h$:$3h = 2a + a\left(\frac{3k}{2a}ight)^2 = 2a + \frac{9k^2}{4a}$.$9k^2 = 4a(3h - 2a) \Rightarrow k^2 = \frac{4a}{3}\left(h - \frac{2a}{3}ight)$.The locus is $y^2 = \frac{4a}{3}\left(x - \frac{2a}{3}ight)$.Comparing with $Y^2 = 4AX$,we have $4A = \frac{4a}{3} \Rightarrow A = \frac{a}{3}$.Vertex is $\left(\frac{2a}{3}, 0ight)$.Focus is $\left(\frac{2a}{3} + A, 0ight) = \left(\frac{2a}{3} + \frac{a}{3}, 0ight) = (a, 0)$.Thus,$(A)$ and $(D)$ are correct.

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