Find the sum of the the intercepts on the axes of coordinates by any tangent to the curve √x +√y=2
Plz solve this problem............
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UP 0 DOWN 0 0 2
2 Answers
21Answer is basically 4 units...
If point is (h,k), m=dy/dx=-(k/h)^(1/2)
X-intercept of tangent turns out as 2(h^{1/2})
Y-intercept = 2(k^1/2)
Adding 2(h^{1/2}+k^{1/2})=4units...
11ya the answer is 4.
it is basically an identity...
that if the curve √x + √y = √a cuts intercepts in the X and Y axes....
the sum of the intercepts is a.
it can be proved in the manner omkar has done...
√x + √y = √a.
differenciating,
(dy/dx)at (h,k) = - √k/h [ (h,k) be a point on tangent]
equation of the tangent:-
-√k/h (x-h) = (y-k).
in this equation, once put x=0 ( for the intercept on x- axis) and y=0 ( for the intercept on y- axis).
then add, the two intercepts... it would sum upto a.
here √a = 2 = √4.
thus the ans is a=4.