A) 27\[{{c}^{3}}\]
B) \[\frac{4}{27}{{c}^{3}}\]
C) \[\frac{27}{4}{{c}^{3}}\]
D) \[\frac{4}{9}{{c}^{3}}\]
Correct Answer: C
Solution :
[c] \[{{x}^{2}}y={{c}^{3}}\] Differentiating w.r.t. x, we have \[{{x}^{2}}\frac{dy}{dx}+2xy=0\] or \[\frac{dy}{dx}=-\frac{2y}{x}\] Equation of the tangent at (h, k) is \[y-k=-\frac{2k}{h}(x-h)\] \[y=0\] gives \[x=\frac{3h}{2}=a\], and x=0 gives\[y=3k=b\]. Now, \[{{a}^{2}}b=\frac{9{{h}^{2}}}{4}3k=\frac{27}{4}{{h}^{2}}k=\frac{27}{4}{{c}^{3.}}\]
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