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.