JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 14)
The tangent $$PT$$ and the normal $$PN$$ to the parabola $${y^2} = 4ax$$ at a point $$P$$ on it meet its axis at points $$T$$ and $$N$$, respectively. The locus of the centroid of the triangle $$PTN$$ is a parabola whose
vertex is $$\left( {{{2a} \over 3},0} \right)$$
directrix is $$x=0$$
latus rectum is $${{{2a} \over 3}}$$
focus is $$(a, 0)$$
Explanation
We have $$G \equiv (h,k)$$.
$$ \Rightarrow h = {{2a + a{t^2}} \over 3},k = {{2at} \over 3}$$
$$ \Rightarrow \left( {{{3h - 2a} \over a}} \right) = {{9{k^2}} \over {4{a^2}}}$$
Therefore, the required parabola is
$${{9{y^2}} \over {4{a^2}}} = {{(3x - 2a)} \over a} = {3 \over a}\left( {x - {{2a} \over 3}} \right)$$
$$ \Rightarrow {y^2} = {{4a} \over 3}\left( {x - {{2a} \over 3}} \right)$$
Hence, the vertex $$ \equiv \left( {{{2a} \over 3},0} \right)$$; focus $$ \equiv (a,0)$$.
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