Applications Of Derivatives Question 104
Question: The area of the triangle formed by the coordinate axes and a tangent to the curve $ xy=a^{2} $ at the point $ (x_1,y_1) $ on it is
[DCE 2001]
Options:
A) $ \frac{a^{2}x_1}{y_1} $
B) $ \frac{a^{2}y_1}{x_1} $
C) $ 2a^{2} $
D) $ 4a^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ y=\frac{a^{2}}{x},,\therefore \frac{dy}{dx}=-\frac{a^{2}}{x^{2}} $
$ \therefore $ At $ (x_1,,y_1), $
$ \frac{dy}{dx}=\frac{-a^{2}}{x_1^{2}} $ Thus tangent to the curve will be $ y-y_1=\frac{-a^{2}}{x_1^{2}}(x-x_1) $
therefore $ yx_1^{2}-y_1x_1^{2}=-a^{2}x+a^{2}x_1 $
therefore $ y’\frac{1}{\sqrt{1-{{( \frac{2x}{1+x^{2}} )}^{2}}}}.\frac{2(1+x^{2})-4x^{2}}{{{(1+x^{2})}^{2}}} $ , $ (\because x_1y_1=a^{2}) $ This meets the x-axis where $ y=0 $
$ \therefore a^{2}x=2a^{2}x_1, $
$ \therefore x=2x_1 $
$ \therefore $ Point on the x-axis is $ (2x_1,0) $ Again tangent meets the y-axis where $ x=0 $
$ \because x_1^{2}y=2a^{2}x_1,,\ \ \ \therefore y=\frac{2a^{2}}{x_1} $ So point on the y-axis is $ ( 0,\frac{2a^{2}}{x_1} ) $ Required area $ =\frac{1}{2}(2x_1)( {{\frac{2a}{x_1}}^{2}} )=2a^{2} $ .
BETA
