Applications-Of-Derivatives Question 454
Question: The sum of intercepts on co-ordinate axes made by tangent to the curve $ \sqrt{x}+\sqrt{y}=\sqrt{a} $ is
[RPET 1999]
Options:
A) $ a $
B) $ 2a $
C) $ 2\sqrt{a} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sqrt{x}+\sqrt{y}=a $ ; $ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0 $ , \ $ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}} $ Hence tangent at (x, y) is $ Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x) $ or $ X\sqrt{y}+Y\sqrt{x}=\sqrt{xy}( \sqrt{x}+\sqrt{y} )=\sqrt{axy} $ or $ \frac{X}{\sqrt{a}\sqrt{x}}+\frac{Y}{\sqrt{a}\sqrt{y}}=1 $ . Clearly its intercepts on the axes are $ \sqrt{a}\sqrt{x} $ and $ \sqrt{a}\sqrt{y} $ . Sum of the intercepts = $ \sqrt{a}( \sqrt{x}+\sqrt{y} )=\sqrt{a}.\sqrt{a}=a $ .
BETA
