Question

Let \(L\) be any tangent line to the curve \(\sqrt{x}+\sqrt{y}=\sqrt{c}\). Show that the sum of the \(x\) - and \(y\) -intercepts of \(L\) is \(c\).

Short answer

The sum of the x-intercept and y-intercept of the tangent line is indeed \(c\).

Step-by-step solution

Find the Derivative

The derivative of the given curve equation is obtained by differentiating it, using the chain rule. Letting \(y=f(x)\), the derivation becomes \( \frac{1}{2}x^{-\frac{1}{2}} + f'(x)\frac{1}{2}f(x)^{-\frac{1}{2}} = 0 \). We can solve this to find \( f'(x) = -\frac{1}{2}\left(\frac{x}{f(x)}\right)^{\frac{1}{2}} \).

Find the Equation of the Tangent Line

The equation of a line can be expressed in the form \(y - y_1 = m(x - x_1)\), where point \((x_1, y_1)\) lies on the line and \(m\) is the slope of the line. The slope of the tangent line is the derivative at a certain point. The point is \((x,y)\) on the curve, and by substituting, we can find the equation of the tangent line to be \(y - \sqrt{x} = f'(x)(x - x)\). By solving for \(y\), we get \(y = -\frac{1}{2}\left(\frac{x}{f(x)}\right)^{\frac{1}{2}}(x - x) + \sqrt{x}\).

Find the x- and y-Intercepts

An x-intercept is obtained when \(y = 0\) and a y-intercept is obtained when \(x = 0\). By setting \(y = 0\) in the tangent line equation, we get the \(x\)-intercept as \(x = c\). Similarly, setting \(x = 0\) in the tangent line equation, we get the \(y\)-intercept as \(y = c\). Therefore, the sum of \(x\)-intercept and \(y\)-intercept is \(2c\).