Standard +0.8 This requires implicit differentiation to find dy/dx, recognizing that vertical tangents occur when dx/dy = 0 (or dy/dx is undefined), then solving the resulting system. It combines multiple techniques and requires conceptual understanding beyond routine differentiation, making it moderately challenging but still within standard C4 scope.
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The diagram shows the curve with equation \(x ^ { 2 } + y ^ { 3 } - 8 x - 12 y = 4\). At each of the points \(P\) and \(Q\) the tangent to the curve is parallel to the \(y\)-axis. Find the coordinates of \(P\) and \(Q\).
\(3y^2-8\frac{dx}{dy}-12\); SC2 for \(\frac{dy}{dx}=\frac{1}{3}(-x^2+8x+12y+4)^{-\frac{2}{3}}\times(-2x+8+12\frac{dy}{dx})\); M1 may be earned for setting correct denominator equal to 0
their \(3y^2\frac{dy}{dx}-12\frac{dy}{dx}=8-2x\) soi; must be two terms on each side and must follow from RHS\(=0\)
M1
their \(2x\frac{dx}{dy}-8\frac{dx}{dy}=-3y^2+12\); must be two terms on each side must follow from RHS\(=0\)
\(\frac{dy}{dx}=\frac{8-2x}{3y^2-12}\) oe
A1
This mark may be implied if \(\frac{dx}{dy}=0\) is substituted and there is no evidence for an incorrect expression for \(\frac{dx}{dy}\)
their \(3y^2-12=0\)
M1*
\(x\neq4\) not required
\(y=(\pm)2\)
A1
A0 if \(\frac{dy}{dx}\) incorrect
Substitution of their positive \(y\) value in original equation
M1dep*
Ignore substitution of \(-2\)
\(x=10\), \(x=-2\) and no others cao
A1
A0 if \(\frac{dy}{dx}\) incorrect; condone omission of formal statement of coordinates \((10,2)\) and \((-2,2)\)
[8]
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3y^2\frac{dy}{dx}$ | B1 | or $2x\frac{dx}{dy}$; if B0B0 M0 |
| $2x-12\frac{dy}{dx}-8$ | B1 | $3y^2-8\frac{dx}{dy}-12$; SC2 for $\frac{dy}{dx}=\frac{1}{3}(-x^2+8x+12y+4)^{-\frac{2}{3}}\times(-2x+8+12\frac{dy}{dx})$; M1 may be earned for setting correct denominator equal to 0 |
| their $3y^2\frac{dy}{dx}-12\frac{dy}{dx}=8-2x$ soi; must be two terms on each side and must follow from RHS$=0$ | M1 | their $2x\frac{dx}{dy}-8\frac{dx}{dy}=-3y^2+12$; must be two terms on each side must follow from RHS$=0$ |
| $\frac{dy}{dx}=\frac{8-2x}{3y^2-12}$ oe | A1 | This mark may be implied if $\frac{dx}{dy}=0$ is substituted and there is no evidence for an incorrect expression for $\frac{dx}{dy}$ |
| their $3y^2-12=0$ | M1* | $x\neq4$ not required |
| $y=(\pm)2$ | A1 | A0 if $\frac{dy}{dx}$ incorrect |
| Substitution of their positive $y$ value in original equation | M1dep* | Ignore substitution of $-2$ |
| $x=10$, $x=-2$ and no others cao | A1 | A0 if $\frac{dy}{dx}$ incorrect; condone omission of formal statement of coordinates $(10,2)$ and $(-2,2)$ |
| **[8]** | | |
6\\
\includegraphics[max width=\textwidth, alt={}, center]{02e31b5d-10dd-42b1-885a-6db610d788c3-2_570_1191_1509_420}
The diagram shows the curve with equation $x ^ { 2 } + y ^ { 3 } - 8 x - 12 y = 4$. At each of the points $P$ and $Q$ the tangent to the curve is parallel to the $y$-axis. Find the coordinates of $P$ and $Q$.
\hfill \mbox{\textit{OCR C4 2014 Q6 [8]}}