Standard +0.8 This requires implicit differentiation of a product-rule-heavy equation, setting dy/dx = 0 for horizontal tangent, then solving the resulting system to prove uniqueness and find the y-coordinate. The algebraic manipulation and proof of uniqueness elevate this above routine stationary point questions, though it follows a standard implicit differentiation framework.
5 The equation of a curve is \(2 x ^ { 2 } y - x y ^ { 2 } = a ^ { 3 }\), where \(a\) is a positive constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis and find the \(y\)-coordinate of this point.
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| State $4xy + 2x^2\dfrac{dy}{dx}$, or equivalent, as derivative of $2x^2y$ | B1 | |
| State $y^2 + 2xy\dfrac{dy}{dx}$, or equivalent, as derivative of $xy^2$ | B1 | |
| Equate attempted derivative of LHS to zero and set $\dfrac{dy}{dx}$ equal to zero (or set numerator equal to zero) | *M1 | $\dfrac{dy}{dx} = \dfrac{y^2 - 4xy}{2x^2 - 2xy}$ |
| Reject $y = 0$ | B1 | Allow from $y^2 - kxy = 0$ |
| Obtain $y = 4x$ | A1 | OE from correct numerator. ISW |
| Obtain an equation in $y$ (or in $x$) and solve for $y$ (or for $x$) in terms of $a$ | DM1 | $8x^3 - 16x^3 = a^3$ or $\dfrac{y^3}{8} - \dfrac{y^3}{4} = a^3$ |
| Obtain $y = -2a$ | A1 | With no errors seen |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Rewrite as $y = \dfrac{a^3}{2x^2 - xy}$ and differentiate | M1 | Correct use of function of a function and implicit differentiation |
| Obtain correct derivative (in any form) | A1 | $\dfrac{dy}{dx} = \dfrac{-a^3\!\left(4x - y - x\dfrac{dy}{dx}\right)}{\left(2x^2 - xy\right)^2}$ |
| Set $\dfrac{dy}{dx}$ equal to zero (or set numerator equal to zero) | *M1 | |
| Obtain $4x - y = 0$ | A1 | |
| Confirm $2x^2 - xy \neq 0$ | B1 | $x = 0$ and $2x = y$ both give $a = 0$ |
| Obtain an equation in $y$ (or in $x$) and solve for $y$ (or for $x$) | DM1 | $8x^3 - 16x^3 = a^3$ or $\dfrac{y^3}{8} - \dfrac{y^3}{4} = a^3$ |
| Obtain $y = -2a$ | A1 | With no errors seen |
5 The equation of a curve is $2 x ^ { 2 } y - x y ^ { 2 } = a ^ { 3 }$, where $a$ is a positive constant. Show that there is only one point on the curve at which the tangent is parallel to the $x$-axis and find the $y$-coordinate of this point.\\
\hfill \mbox{\textit{CAIE P3 2019 Q5 [7]}}