JEE MAIN - Mathematics (2022 - 25th June Morning Shift - No. 14)
Let $$x = 2t$$, $$y = {{{t^2}} \over 3}$$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $$SA \bot BA$$, where A is any point on the conic. If k is the ordinate of the centroid of the $$\Delta$$SAB, then $$\mathop {\lim }\limits_{t \to 1} k$$ is equal to :
$${{17} \over {18}}$$
$${{19} \over {18}}$$
$${{11} \over {18}}$$
$${{13} \over {18}}$$
Explanation
$$x = 2t,\,y = {2 \over 3}$$
$$t \to 1\,\,\,A \equiv \left( {2,{1 \over 3}} \right)$$
Given conic is $${x^2} = 12y \Rightarrow S \equiv (0,3)$$
Let $$B \equiv (0,\beta )$$
Given $$SA\, \bot \,BA$$
$$\left( {{{{1 \over 3}} \over {2 - 3}}} \right)\left( {{{\beta - {1 \over 3}} \over { - 2}}} \right) = - 1$$
$$ \Rightarrow \left( {\beta - {1 \over 3}} \right){1 \over 3} = - 2$$
$$ \Rightarrow \beta = {1 \over 3}\left( {{{ - 17} \over 3}} \right)$$
$$\mathop {Ordinate\,of\,centroid}\limits_{(as\,t \to 1)} = k = {{\beta + {1 \over 3} + 3} \over 3}$$
$$ = {{{{ - 17} \over 9} + {{10} \over 3}} \over 3} = {{13} \over {18}}$$
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