A) 2
B) \[4/3\]
C) \[1/2\]
D) \[3/4\]
Correct Answer: B
Solution :
\[\frac{{{x}^{2}}}{\alpha }+\frac{{{y}^{2}}}{4}=1\Rightarrow \frac{2x}{\alpha }+\frac{2y}{4},\frac{dy}{dx}=0\] \[\Rightarrow \frac{dy}{dx}=\frac{-4x}{\alpha y}\] ?(i) \[{{y}^{3}}=16x\Rightarrow 3{{y}^{2}}.\frac{dy}{dx}=16\Rightarrow \frac{dy}{dx}=\frac{16}{3{{y}^{2}}}...\](ii) Since curves intersects at right angles \[\therefore \]\[\frac{-4x}{\alpha y}\times \frac{16}{3{{y}^{2}}}=-1\Rightarrow 3\alpha {{y}^{3}}=64x\] \[\Rightarrow \]\[\alpha =\frac{64x}{3\times 16x}=\frac{4}{3}\]
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