Find the equation of the tangent to the curve \(\sqrt{x}+\sqrt{y}=a\)

1.	Find the equation of the tangent to the curve \(\sqrt{x}+\sqrt{y}=a\), at the point (a2/4, a2/4).
Answer» finding slope of the tangent by differentiating the curve \(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}(\frac{dy}{dx})=0\) \(\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{y}}\) at (\(\frac{a^2}{4},\frac{a^2}{4}\)) slope m, is – 1 the equation of the tangent is given by y – y₁ = m(x – x₁) \(y-\frac{a^2}{4}\)\(-1(x-\frac{a^2}{4})\) \(x+y=\frac{a^2}{2}\)

Find the equation of the tangent to the curve \(\sqrt{x}+\sqrt{y}=a\), at the point (a2/4, a2/4).

Answer»

finding slope of the tangent by differentiating the curve

\(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}(\frac{dy}{dx})=0\)

\(\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{y}}\)

at (\(\frac{a^2}{4},\frac{a^2}{4}\)) slope m, is – 1

the equation of the tangent is given by y – y₁ = m(x – x₁)

\(y-\frac{a^2}{4}\)\(-1(x-\frac{a^2}{4})\)

\(x+y=\frac{a^2}{2}\)