A) 8a/27
B) 27/8b
C) 8b/27
D) 8/27
Correct Answer: C
Solution :
[c] Here, \[b{{y}^{2}}={{(x+a)}^{3}}\] Differentiating both the sides, we get \[2by\frac{dy}{dx}=3{{(x+a)}^{2}}\Rightarrow \frac{dy}{dx}=\frac{3{{(x+a)}^{2}}}{2by}\] \[\therefore \] length of subnormal \[SN=y\frac{dy}{dx}=\frac{3}{2}\frac{{{(x+a)}^{2}}}{b}\therefore \] length of subtangent \[ST=y.\frac{dx}{dy}=\frac{2b{{y}^{2}}}{3{{(x+a)}^{2}}}\therefore p(SN)=q{{(ST)}^{2}}\] \[\Rightarrow \frac{p}{q}=\frac{{{(ST)}^{2}}}{(SN)}=\frac{8}{27}\frac{{{b}^{3}}{{y}^{4}}}{{{(x+a)}^{6}}}=\frac{8b}{27}\] \[\left( \because \frac{{{b}^{2}}{{y}^{4}}}{{{(x+a)}^{6}}}=1 \right)\]
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