Question

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

Short answer

The sum of x- and y-intercepts of the tangent to the curve \(\sqrt x + \sqrt y = \sqrt c \) is equal to c.

Step-by-step solution

Differentiate the equation of the curve

Differentiate the curve equation \(\sqrt x + \sqrt y = \sqrt c \)with respect to \(x\)on both sides.

\(\begin{aligned}\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt y }}\left( {y'} \right) &= 0\\y' &= - \sqrt {\frac{y}{x}} \end{aligned}\)

The slope of the tangent to the curve is \( - \sqrt {\frac{y}{x}} \).

Find the equation of the tangent 

The equation of the tangent at the point \(\left( {{x_0},{y_0}} \right)\) is:

\(\begin{aligned}y - {y_0} &= - \sqrt {\frac{{{y_0}}}{{{x_0}}}} \left( {x - {x_0}} \right)\\y &= {y_0} - \sqrt {\frac{{{y_0}}}{{{x_0}}}} \left( {x - {x_0}} \right)\end{aligned}\)

Find x- and y-intercept of the tangents

Substitute 0 for \(x\) to find the value of the y-intercept.

\(\begin{aligned}y &= {y_0} - \sqrt {\frac{{{y_0}}}{{{x_0}}}} \left( {0 - {x_0}} \right)\\ &= {y_0} + \sqrt {{y_0}{x_0}} \end{aligned}\)

Substitute 0 for y to find the value of x-intercept.

\(\begin{aligned} 0 &= {y_0} - \sqrt {\frac{{{y_0}}}{{{x_0}}}} \left( {x - {x_0}} \right)\\x &= {x_0} + \sqrt {{y_0}{x_0}} \end{aligned}\)

Find the sum of x- and y-intercepts

Obtain the sum of x and y-intercepts.

\(\begin{aligned} x\;{\rm{intercept}} + y\,{\rm{intercept}} &= {x_0} + \sqrt {{y_0}{x_0}} + {y_0} + \sqrt {{y_0}{x_0}} \\ &= {x_0} + {y_0} + 2\sqrt {{y_0}{x_0}} \\ &= {\left( {\sqrt {{x_0}} + \sqrt {{y_0}} } \right)^2}\\ &= {\left( {\sqrt c } \right)^2}\\& = c\end{aligned} \)

Thus, the sum of x- and y-intercepts of the tangent to the curve \(\sqrt x + \sqrt y = \sqrt c \) is equal to c.