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 the \(x\)and \(y\)intercepts of tangent line to the curve \(\sqrt x + \sqrt y = \sqrt c \)is \(c.\)

Step-by-step solution

Step 1:Given Information.

The given expression is\(\sqrt x + \sqrt y = \sqrt c .\)

Meaning of Tangent line

The line that touches the curve at a point called the point of tangency is a tangent line.

Show that the sum of the\((a,b)\)and \(y\)intercepts.

Differentiate throughout with respect to \(x\)

\(\begin{array}{l}\frac{{d(\sqrt x )}}{{dx}} + \frac{{d(\sqrt y )}}{{dx}} = \frac{{d(\sqrt c )}}{{dx}}\\\frac{1}{{2\sqrt x }} + \underbrace {\frac{{d(\sqrt y )}}{{dy}} \cdot \frac{{dy}}{{dx}}}_{{\rm{CHAIN RULE }}} = 0\\\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt y }} \cdot \frac{{dy}}{{dx}} = 0\\\frac{1}{{2\sqrt y }} \cdot \frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x }}\end{array}\)

Further solve

\(\begin{array}{l}\frac{{dy}}{{dx}} = - \frac{{2\sqrt y }}{{2\sqrt x }}\\\frac{{dy}}{{dx}} = - \frac{{\sqrt y }}{{\sqrt x }}\end{array}\)

Let \((a,b)\)be a point on the curve.

Find the slope of the tangent.

\({\left. {\frac{{dy}}{{dx}}} \right|_{(a,b)}} = - \frac{{\sqrt b }}{{\sqrt a }}\)

Find the equation of the tangent.

\(\frac{{y - b}}{{x - a}} = - \frac{{\sqrt b }}{{\sqrt a }}\)

Cross Multiply

\(\begin{array}{l}y\sqrt a - b\sqrt a = - x\sqrt b + a\sqrt b \\x\sqrt b + y\sqrt a = a\sqrt b + b\sqrt a \\\frac{{x\sqrt b }}{{a\sqrt b + b\sqrt a }} + \frac{{y\sqrt a }}{{a\sqrt b + b\sqrt a }} = 1\\\frac{x}{{a + \sqrt {ab} }} + \frac{y}{{\sqrt {ab} + b}} = 1\end{array}\)

Find the sum of\(x\),and \(y\)intercepts.

\(\begin{array}{c}a + 2\sqrt {ab} + b = {(\sqrt a )^2} + 2\sqrt a \sqrt b + {(\sqrt b )^2}\\ = {(\sqrt a + \sqrt b )^2}\\ = {(\sqrt c )^2}\\ = c\end{array}\)