A curve C has equation x² + 4xy – y² + 20 = 0. Show that, at stationary points on C, x = −2y. Find the coordinates of the stationary points on C, and determine their nature by considering the value of d²y/dx² at the stationary points.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

1. [M1A1] Differentiating implicitly wrt x: 2x+4(xy'+ y)-2yy' = 0 (*)
2. [A1] y'=0⇒2x+4y=0⇒x=−2y (AG)
3. [M1A1] At stationary points (y'=0), substitute x=-2y into the original equation: (-2y)² + 4(-2y)y - y² + 20 = 0 => 4y² - 8y² - y² + 20 = 0 => -5y² = -20 => 5y² = 20 => y² = 4 => y = ±2
4. [A1] Coordinates of stationary points are (4, -2) and (-4,2)
5. [M1A1] Differentiating (*) again wrt x: 1+2(xy'' + y' + y')−(yy'' +[y']²) = 0 (or by quotient rule)
6. [M1A1] At (4,-2) with y' = 0 : 1+2(4y'' + 0 + 0) - (-2y'' + 0) = 0 => 1+8y''+2y'' = 0 => 1+10y''=0 => y'' = -1/10 => maximum.
7. [A1] At (-4, 2) with y' = 0 : 1+2(-4y'' + 0 + 0) - (2y'' + 0) = 0 => 1-8y''-2y'' = 0 => 1-10y''=0 => y'' = 1/10 => minimum.