Question

Let $x=2 t, y=\frac{t^{2}}{3}$ be a conic. Let $S$ be the focus and $B$ be the point on the axis of the conic such that $SA \perp BA$, where $A$ is any point on the conic. If $k$ is the ordinate of the centroid of $\Delta SAB$, then $\displaystyle\lim _{ t \rightarrow 1} k$ is equal to

• A. $\frac{17}{18}$
• B. $\frac{19}{18}$
• C. $\frac{11}{18}$
• D. $\frac{13}{18}$

Answer 1

Answer: (D)

parabola $x ^{2}=12 y$
$SA \perp SB$
so, $m _{ AS } \cdot m _{ AB }=-1$
$\frac{\left(3-\frac{t^{2}}{3}\right)}{(0-2 t)} \cdot \frac{\left(\alpha-\frac{t^{2}}{3}\right)}{(0-2 t)}=-1$
by solving
$3 \alpha=\frac{27 t^{2}+t^{4}}{t^{2}-9}$
ordinate of centriod of $\Delta SAB = K =\frac{\alpha+\frac{ t ^{2}}{3}+3}{3}$
$k =\frac{9+3 \alpha+ t ^{2}}{9}$
$\displaystyle\lim _{t \rightarrow 1} k=\displaystyle\lim _{t \rightarrow 1} \frac{1}{9}\left(9+t^{2}+\frac{27 t^{2}+t^{4}}{\left(t^{2}-9\right)}\right)=\frac{13}{18}$