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Mathematics Question on Conic sections

Let x=2t, y=t23x=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⊥BA, where A is any point on the conic. If k is the ordinate of the centroid of the ΔSAB, then lim⁡t→1k\lim\limits_{t \to 1} k is equal to

A

1718\frac {17}{18}

B

1918\frac {19}{18}

C

1118\frac {11}{18}

D

1318\frac {13}{18}

Answer

1318\frac {13}{18}

Explanation

Solution

x=2t, y=23x=2t,\ y=\frac 23
For t=1t=1
A=(2,13)A=(2,\frac 13)
Conic is x^2 = 12y $$⇒ S = (0, 3)
Let B=(0,β)B = (0, β)
Given SA⊥BASA⊥BA
(132−3)(β−13−2)=−1(\frac {\frac 13}{2−3})(\frac {β−\frac 13}{−2})=−1

(β−13)13=−2(β−\frac 13)\frac 13=−2
β=13(−173)β=\frac 13(−\frac {17}{3})
Ordinate of centroid,
K=β+13+33K = \frac {β+\frac 13+3}{3}

=−179+1033=\frac {−\frac {17}{9}+\frac {10}{3}}{3}

=1318= \frac {13}{18}

So, the correct option is (D): 1318\frac {13}{18}