Challenging +1.2 This is a multi-step implicit differentiation problem requiring: (1) implicit differentiation of a moderately complex expression including exponential terms, (2) finding dy/dx at a specific point, (3) finding equations of tangent and normal lines, (4) finding y-intercepts, and (5) calculating triangle area. While it involves several techniques and careful algebra, each individual step follows standard procedures taught in A-level Further Maths, making it moderately above average difficulty but not requiring novel insight.
5 The curve \(C\) has equation
$$3 x ^ { 2 } - 5 x y + \mathrm { e } ^ { 2 y - 4 } + 6 = 0$$
The point \(P\) with coordinates \(( 1,2 )\) lies on \(C\). The tangent to \(C\) at \(P\) meets the \(y\)-axis at the point \(A\) and the normal to \(C\) at \(P\) meets the \(y\)-axis at the point \(B\).
Find the exact area of triangle \(A B P\).
Use equation of normal with their \(m_N = -\frac{1}{m_T}\), the point \((1, 2)\) and \(x = 0\)
For reference: \(y = \frac{5}{4}\)
Area of triangle \(= \frac{1}{2}(1)\left(\frac{10}{3} - \frac{5}{4}\right)\)
M1
Area must be \(\frac{1}{2}(1)
\(= \frac{25}{24}\)
A1
[8 marks total]
$\frac{d}{dx}\left(e^{2y-4}\right) = 2e^{2y-4}\frac{dy}{dx}$ | B1 | Correct application of implicit differentiation
$\frac{d}{dx}(5xy) = 5y + 5x\frac{dy}{dx}$ | M1 | Attempt at product rule (two terms, + sign)
$6x - 5y - 5x\frac{dy}{dx} + 2e^{2y-4}\frac{dy}{dx} = 0$ | A1 |
$6 - 10 - 5\frac{dy}{dx} + 2\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \ldots$ | M1 | Substitute for $x = 1$ and $y = 2$ and attempt to find $\frac{dy}{dx}$
| | For reference: $\frac{dy}{dx} = -\frac{4}{3}$
Tangent: $y - 2 = -\frac{4}{3}(x - 1)$ with $x = 0$ | M1 | Use equation of tangent with their $m_T$, the point $(1, 2)$ and $x = 0$
| | For reference: $y = \frac{10}{3}$
Normal: $y - 2 = \frac{3}{4}(x - 1)$ with $x = 0$ | M1 | Use equation of normal with their $m_N = -\frac{1}{m_T}$, the point $(1, 2)$ and $x = 0$
| | For reference: $y = \frac{5}{4}$
Area of triangle $= \frac{1}{2}(1)\left(\frac{10}{3} - \frac{5}{4}\right)$ | M1 | Area must be $\frac{1}{2}(1)|y_T - y_N|$ where $y_N$ is the y-value of the normal at $x = 0$ and $y_T$ is the corresponding value for the tangent – dependent on all previous M marks
$= \frac{25}{24}$ | A1 |
**[8 marks total]**
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5 The curve $C$ has equation
$$3 x ^ { 2 } - 5 x y + \mathrm { e } ^ { 2 y - 4 } + 6 = 0$$
The point $P$ with coordinates $( 1,2 )$ lies on $C$. The tangent to $C$ at $P$ meets the $y$-axis at the point $A$ and the normal to $C$ at $P$ meets the $y$-axis at the point $B$.
Find the exact area of triangle $A B P$.
\hfill \mbox{\textit{OCR H240/03 2018 Q5 [8]}}