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Question
Find the equation of the tangent to the curve \[\sqrt{x} + \sqrt{y} = a\] at the point \[\left( \frac{a^2}{4}, \frac{a^2}{4} \right)\] ?
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Solution
\[\sqrt{x} + \sqrt{y} = a\]
\[\text { Differentiating both sides w.r.t.x}, \]
\[ \Rightarrow \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sqrt{y}}{\sqrt{x}}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( \frac{a^2}{4}, \frac{a^2}{4} \right)\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( \frac{a^2}{4}, \frac{a^2}{4} \right) =\frac{- \sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}=-1\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{a^2}{4} = - 1\left( x - \frac{a^2}{4} \right)\]
\[ \Rightarrow y - \frac{a^2}{4} = - x + \frac{a^2}{4}\]
\[ \Rightarrow x + y = \frac{a^2}{2}\]
