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Q.
The point on the curve $ \sqrt{x}+\sqrt{y}=\sqrt{a} $ , the normal at which is parallel to the $ x- $ axis, is:
Jharkhand CECEJharkhand CECE 2005
Solution:
The normal is parallel to $ x- $ axis, if $ \frac{dx}{dy}=0 $
Given equation of curve is $ \sqrt{x}+\sqrt{y}=\sqrt{a} $ ... (i)
On differentiating w.r.t. $ x, $
we get $ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}=\frac{dy}{dx}=0 $
$ \Rightarrow $ $ \frac{dx}{dy}=\sqrt{\frac{x}{y}} $
Since, the normal is parallel to $ x- $ axis.
$ \therefore \frac{dx}{dy}=0 $
$ \Rightarrow -\sqrt{\frac{x}{y}}=0 $
$ \Rightarrow x=0 $
$ \therefore $ From Eq. (i) $ y=a $
$ \therefore $ Required point is $ (0,\,\,a) $ .