A) 2
B) 1
C) 0
D) 4
Correct Answer: B
Solution :
[b] Given curve is \[{{x}^{3/2}}+{{y}^{3/2}}+{{y}^{3/2}}=2{{a}^{3/2}}...(1)\] \[\therefore \frac{3}{2}\sqrt{x}+\frac{3}{2}\sqrt{y}\frac{dy}{dx}=0\] or \[\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{y}}\] Since the tangent is equally inclined to axes, \[\frac{dy}{dx}=\pm 1\therefore -\frac{\sqrt{x}}{\sqrt{y}}=\pm 1or-\frac{\sqrt{x}}{\sqrt{y}}=-1\] \[\therefore \sqrt{x}=\sqrt{y}[\because \sqrt{x}>0,\sqrt{y}>0]\] Putting \[\sqrt{y}=\sqrt{x}\] in (1), we get \[2{{x}^{3/2}}=2{{a}^{3/2}}\,\,\,or\,\,\,{{x}^{3}}={{a}^{3}}\]. Therefore, \[x=a\] and, so, \[y=a\].
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