Date | November 2015 | Marks available | 4 | Reference code | 15N.1.hl.TZ0.7 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
A curve is defined by \(xy = {y^2} + 4\).
Show that there is no point where the tangent to the curve is horizontal.
Find the coordinates of the points where the tangent to the curve is vertical.
Markscheme
\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1
a horizontal tangent occurs if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) so \(y = 0\) M1
we can see from the equation of the curve that this solution is not possible \((0 = 4)\) and so there is not a horizontal tangent R1
[4 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y - x}}\) or equivalent with \(\frac{{{\text{d}}x}}{{{\text{d}}y}}\)
the tangent is vertical when \(2y = x\) M1
substitute into the equation to give \(2{y^2} = {y^2} + 4\) M1
\(y = \pm 2\) A1
coordinates are \((4,{\text{ }}2),{\text{ }}( - 4,{\text{ }} - 2)\) A1
[4 marks]
Total [8 marks]