The tangent to the curve ` sqrt(x)+sqrt(y)=sqrt(a)` at any poin on it cuts the axes Ox and Oy at P and Q respectively then `OP+OQ` is

Options

A. 2a
B. a
C. `(a)/(2)`
D. none of these

Answer

B

Explanation

Let P`(x_(1),y_(1))` be a point on the curve ` sqrt(x)+sqrt(y)=sqrt(a). `
Then,
` sqrt(x_(1))+sqrt(y_(1))=sqrt(a) " "...(i)`
Now,
` sqrt(x)+sqrt(y)=sqrt(a)`
` rArr (1)/(2sqrt(x))+(1)/(2sqrt(y)) (dy)/(dx)=0 rArr (dy)/(dx) = -(sqrt(y))/(sqrt(x)) rArr ((dy)/(dx))_(P)=-sqrt((y_(1))/(x_(1)))`
The equation of the tangent to the given curve at point `P(x_(1),y_(1))` is
` y-y_(1)=-sqrt((y_(1))/(x_(1)))(x-x_(1)) `
` rArr (x)/(sqrt(x_(1)))+(y)/(sqrt(y_(1))) = sqrt(x_(1))+ sqrt(y_(1)) rArr (x)/(sqrt(x_(1))) + (y)/(sqrt(y_(1)))=sqrt(a) `
[Using (i)]
This cuts the coordinate axes at `A(sqrt(ax_(1)),0)` and `B(0, sqrt(ay_(1)))`
` therefore OA+OB= sqrt(ax_(1))+sqrt(ay_(1))=sqrt(a)(sqrt(x_(1))+sqrt(y_(1)))= sqrt(a) xx sqrt(a)=a. `