Challenging +1.2 This question requires converting a Cartesian equation to polar form (r² = 2cos(2θ)), finding the maximum value of r by differentiating with respect to θ, and converting back to Cartesian coordinates. While it involves multiple steps and polar coordinate manipulation, the techniques are standard for Further Maths P3 students and the differentiation is straightforward once in polar form.
6
\includegraphics[max width=\textwidth, alt={}, center]{b2136f5d-0d66-4524-bb76-fcc4cb59150c-3_405_914_260_612}
The diagram shows the curve \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2 \left( x ^ { 2 } - y ^ { 2 } \right)\) and one of its maximum points \(M\). Find the coordinates of \(M\).
Set \(\frac{dy}{dx}\) equal to zero and obtain a horizontal equation
M1
Obtain a correct equation, e.g. \(x^2 + y^2 = 1\), from correct work
A1
By substitution in the curve equation, or otherwise, obtain an equation in \(x^2\) or \(y^2\)
M1
Obtain \(x = \frac{1}{2}\sqrt{3}\)
A1
Obtain \(y = \frac{1}{2}\)
A1
7 marks
Obtain correct derivative of RHS in any form | B1 |
Obtain correct derivative of LHS in any form | B1 |
Set $\frac{dy}{dx}$ equal to zero and obtain a horizontal equation | M1 |
Obtain a correct equation, e.g. $x^2 + y^2 = 1$, from correct work | A1 |
By substitution in the curve equation, or otherwise, obtain an equation in $x^2$ or $y^2$ | M1 |
Obtain $x = \frac{1}{2}\sqrt{3}$ | A1 |
Obtain $y = \frac{1}{2}$ | A1 | 7 marks
6\\
\includegraphics[max width=\textwidth, alt={}, center]{b2136f5d-0d66-4524-bb76-fcc4cb59150c-3_405_914_260_612}
The diagram shows the curve $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2 \left( x ^ { 2 } - y ^ { 2 } \right)$ and one of its maximum points $M$. Find the coordinates of $M$.
\hfill \mbox{\textit{CAIE P3 2014 Q6 [7]}}