If the tangent to the curve √x+√y=√a at any point on it cuts the axes OX a

Question -

If the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at any point on it cuts the axes OX and OY at P and Q respectively, then OP +OQ is

Options:

A .

\frac{a}{2}

B . a

C . 2a

D . 4a

Answer: Option B
:
B

\sqrt{x} + \sqrt{y} = \sqrt{a} . . . . . (i)

\Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0

∴ \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}}

Equation of tangent at

(x_{1} y_{1}) i s y - y_{1} = - \frac{\sqrt{y_{1}}}{\sqrt{x_{1}}} (x - x_{1})

\Rightarrow \frac{x}{\sqrt{x_{1}}} + \frac{y}{\sqrt{y_{1}}} = \sqrt{a}; \Rightarrow o p = \sqrt{a} \sqrt{x_{1}}, O Q = \sqrt{a} \sqrt{y_{1}} ∴ O P + O Q = \sqrt{a}

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