The point P, with coordinates \((p,{\text{ }}q)\) , lies on the graph of \({x^{\frac{1}{2}}} + {y^{\frac{1}{2}}} = {a^{\frac{1}{2}}}\) , \(a > 0\) .

The tangent to the curve at P cuts the axes at (0, m) and (n, 0) . Show that m + n = a .

\({x^{\frac{1}{2}}} + {y^{\frac{1}{2}}} = {a^{\frac{1}{2}}}\)

\(\frac{1}{2}{x^{ - \frac{1}{2}}} + \frac{1}{2}{y^{ - \frac{1}{2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{{\frac{1}{{2\sqrt x }}}}{{\frac{1}{{2\sqrt y }}}} = - \sqrt {\frac{y}{x}} \)     A1

Note: Accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 - \frac{{{a^{\frac{1}{2}}}}}{{{x^{\frac{1}{2}}}}}\) from making y the subject of the equation, and all correct subsequent working

therefore the gradient at the point P is given by

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \sqrt {\frac{q}{p}} \)     A1

equation of tangent is \(y - q = - \sqrt {\frac{q}{p}} (x - p)\)     M1

\((y = - \sqrt {\frac{q}{p}} x + q + \sqrt q \sqrt p )\)

x-intercept: y = 0, \(n = \frac{{q\sqrt p }}{{\sqrt q }} + p = \sqrt q \sqrt p  + p\)     A1

y-intercept: x = 0, \(m = \sqrt q \sqrt p  + q\)     A1

\(n + m = \sqrt q \sqrt p  + p + \sqrt q \sqrt p  + q\)     M1

\( = 2\sqrt q \sqrt p  + p + q\)

\( = {\left( {\sqrt p  + \sqrt q } \right)^2}\)     A1

\( = a\)     AG

[8 marks]

Many candidates were able to perform the implicit differentiation. Few gained any further marks.