Shortest distance between the curves $9 x^{2} + 9 y^{2} - 30 y + 16 = 0 a n d y^{2} = x^{3}$ is

\frac{\sqrt{13}}{3} - 1

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\frac{\sqrt{13}}{3} + 1

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\frac{\sqrt{13}}{3} + 2

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Solution

The correct option is A $\frac{\sqrt{13}}{3} - 1$
$E q u a t i o n o f n o r m a l t o y^{2} = x^{3} a t a n y p o i n t (t^{2}, t^{3}) i s y - t^{3} = \frac{- 2}{3 t} (x - t^{2}) w i l l p a s s e s t h r o u g h [0, \frac{15}{9}] i . e ., c e n t r e o f t h e c i r c l e \frac{5}{3} - t^{3} = - \frac{2}{3 t} (- t^{2})$
$3 t^{3} + 2 t - 5 = 0 \Rightarrow t = 1$
$\Rightarrow$ point corresponding to shortest distance (1, 1)
$s h o r t e s t d i s t a n c e = d i s t a n c e f r o m (1, 1) t o c e n t r e o f c i r c l e r a d i u s$
$= \sqrt{1 + \frac{4}{9}} - 1 = \frac{\sqrt{13}}{3} - 1$

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