Differentiation & Applications: Implicit — NYJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
The equation of a curve \(C\) is \(x^3+xy+2y^3=k\), where \(k\) is a constant.
- Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\).
It is given that \(C\) has a tangent which is parallel to the \(y\)-axis.
- Show that the \(y\)-coordinate of the points of contact of the tangent with \(C\) must satisfy \[ 216y^6+4y^3+k=0. \] Hence show that \(k\leq\dfrac{1}{54}\).
- Find the possible values of \(k\) in the case where the line \(x=-6\) is a tangent to \(C\).
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(a)
Differentiating \(x^3+xy+2y^3=k\) with respect to \(x\): \[ 3x^2+x\frac{\mathrm{d}y}{\mathrm{d}x}+y+6y^2\frac{\mathrm{d}y}{\mathrm{d}x}=0 \]
(b)
If the tangent is parallel to the \(y\)-axis, then \(x+6y^2=0\), i.e. \(x=-6y^2\).
The point of contact of the tangent with \(C\) satisfies: \[\begin{aligned} (-6y^2)^3+(-6y^2)(y)+2y^3 &= k\\ -216y^6-6y^3+2y^3 &= k\\ -216y^6-4y^3 &= k \end{aligned}\] \[ \therefore\quad 216y^6+4y^3+k=0 \quad\text{(shown)} \]
Treating this as a quadratic in \(y^3\): for \(y\) to be real, the discriminant must be \(\geq 0\): \[\begin{aligned} 4^2-4(216)(k) &\geq 0\\ 4 &\geq 216k\\ k &\leq \frac{4}{216}=\frac{1}{54} \quad\text{(shown)} \end{aligned}\]
(c)
When \(x=-6\): \[\begin{aligned} x = -6y^2 &\implies -6y^2 = -6\\ &\implies y = \pm 1 \end{aligned}\]
For \(y=1\): \(k=(-6)^3+(-6)(1)+2(1)^3=-216-6+2=-220\).
For \(y=-1\): \(k=(-6)^3+(-6)(-1)+2(-1)^3=-216+6-2=-212\).