Question

Let $x=2t,y=\frac{t^2}{3}$ 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 $\underset{t\to 1}{\lim }k$ is equal to

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

Answer 1

Answer: (D)

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