The equation of the tangent to the curve $\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$ at the point $(x_{1}, y_{1})$ is $\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$. Then, the value of $k$ is

Given curve is

$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$ ... (i)

On differentiating w.r.t. to $x$, we get

$\dfrac {1}{\sqrt {a}} \cdot \dfrac {1}{2\sqrt {x}} + \dfrac {1}{\sqrt {b}} \cdot \dfrac {1}{2\sqrt {y}} \dfrac {dy}{dx} = 0$

$\Rightarrow \dfrac {dy}{dx} = -\dfrac {\sqrt {b}\sqrt {y}}{\sqrt {a}\sqrt {x}} \Rightarrow \left [\dfrac {dy}{dx}\right ]_{(x_{1}, y_{1})} = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}}$

Equation of tangent passing through the point $(x_{1}, y_{1})$ is

$(y - y_{1}) = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}} (x - x_{1})$

$\dfrac {y}{\sqrt {by_{1}}} - \dfrac {y_{1}}{\sqrt {by_{1}}} = -\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {x_{1}}{\sqrt {ax_{1}}}$

$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = \dfrac {x_{1}}{\sqrt {ax_{1}}} + \dfrac {y_{1}}{\sqrt {by_{1}}} = \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}}$

$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = 1$ [from Eq. (i)]

$\left [\because at (x_{1}, y_{1}), \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}} = 1\right ]$

But $\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$ (given)

Therefore, $k = 1$