Question

Find the equation of the tangent to the curve \[\sqrt{x} + \sqrt{y} = a\] at the point \[\left( \frac{a^2}{4}, \frac{a^2}{4} \right)\] ?

Answer 1

\[\sqrt{x} + \sqrt{y} = a\]

\[\text { Differentiating both sides w.r.t.x}, \]

\[ \Rightarrow \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- \sqrt{y}}{\sqrt{x}}\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( \frac{a^2}{4}, \frac{a^2}{4} \right)\]

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( \frac{a^2}{4}, \frac{a^2}{4} \right) =\frac{- \sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}=-1\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{a^2}{4} = - 1\left( x - \frac{a^2}{4} \right)\]

\[ \Rightarrow y - \frac{a^2}{4} = - x + \frac{a^2}{4}\]

\[ \Rightarrow x + y = \frac{a^2}{2}\]