The sum of the intercepts made on the axes of coordinates by any tangent to the curve x+y=a is equal to

A
$2 a$
B
$a$
C
$\frac{a}{2}$
D
none of these

Solution:

Let $P (x_{1}, y_{1})$ be a point on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$

Then,

$\sqrt{x_{1}} + \sqrt{y_{1}} = \sqrt{a}$

Now,

$\begin{array}{l} \sqrt{x} + \sqrt{y} = \sqrt{a} \\ \Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{p} = - \sqrt{\frac{y_{1}}{x_{1}}} \end{array}$

The equation of the tangent to the given curve at point $P (x_{1}, y_{1})$ is

$\begin{array}{l} y - y_{1} = - \sqrt{\frac{y_{1}}{x_{1}}} (x - x_{1}) \\ \Rightarrow \frac{x}{\sqrt{x_{1}}} + \frac{y}{\sqrt{y_{1}}} = \sqrt{x_{1}} + \sqrt{y_{1}} \Rightarrow \frac{x}{\sqrt{x_{1}}} + \frac{y}{\sqrt{y_{1}}} = \sqrt{a} \end{array}$

Using (i)]
This cuts the coordinate axes at $A (\sqrt{a x_{1}}, 0) and B (0, \sqrt{a y_{1}})$

$∴ O A + O B = \sqrt{a x_{1}} + \sqrt{a y_{1}} = \sqrt{a} (\sqrt{x_{1}} + \sqrt{y_{1}}) = \sqrt{a} \times \sqrt{a} = a .$

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is equal to

Solution:

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