IB Mathematics AHL 3.14 Vector equation of a line in planes AA SL Paper 3
Question 1
(a) Topic-SL 2.10 Solving equations, both graphically and analytically.
(b) Topic-SL 5.8 Points of inflexion with zero and non-zero gradients.
(c) Topic-SL 5.1 Introduction to the concept of a limit.
(d) Topic-AHL 5.14 Implicit differentiation.
(e) Topic-SL 5.6 The product and quotient rules.
(f) Topic-SL 5.4 Tangents and normals at a given point, and their equations
(g) Topic-AHL 3.14 Vector equation of a line in two and three.
dimensions:
This question asks you to explore properties of a family of curves of the type \(y^2 = x^3 + ax + b\) for various values of \(a\) and \(b\), where \(a, b \in \mathbb{N}\).
(a) On the same set of axes, sketch the following curves for \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\), clearly indicating any points of intersection with the coordinate axes.
(i) \(y^2 = x^3\), \(x \geq 0\)
(ii) \(y^2 = x^3 + 1\), \(x \geq -1\)
(b)
(i) Write down the coordinates of the two points of inflexion on the curve \(y^2 = x^3 + 1\).
(ii) By considering each curve from part (a), identify two key features that would distinguish one curve from the other.
Now, consider curves of the form \(y^2 = x^3 + b\), for \(x \geq \sqrt[3]{b}\), where \(b \in \mathbb{Z}^+\).
(c) By varying the value of \(b\), suggest two key features common to these curves.
Next, consider the curve \(y^2 = x^3 + x\), \(x \geq 0\).
(d)
(i) Show that \(\frac{dy}{dx} = \pm \frac{3x^2 + 1}{2\sqrt{x^3 + x}}, \, \text{for} \, x > 0.\)
(ii) Hence deduce that the curve \(y^2 = x^3 + x\) has no local minimum or maximum points.
The curve \(y^2 = x^3 + x\) has two points of inflexion. Due to the symmetry of the curve, these points have the same \(x\)-coordinate.
(e) Find the value of this \(x\)-coordinate, giving your answer in the form \(x = \frac{p\sqrt{3} + q}{r}\), where \(p, q, r \in \mathbb{Z}\).
\( P(x, y) \) is defined to be a rational point on a curve if \( x \) and \( y \) are rational numbers.
The tangent to the curve \( y^2 = x^3 + ax + b \) at a rational point \( P \) intersects the curve at another rational point \( Q \).
Let \( C \) be the curve \( y^2 = x^3 + 2 \), for \( x \geq -\sqrt[3]{2} \). The rational point \( P(-1, -1) \) lies on \( C \).
(f)
(i) Find the equation of the tangent to \( C \) at \( P \).
(ii) Hence, find the coordinates of the rational point \( Q \) where this tangent intersects \( C \), expressing each coordinate as a fraction.
(g) The point \( S(-1, 1) \) also lies on \( C \). The line \([QS]\) intersects \( C \) at a further point. Determine the coordinates of this point.
▶️Answer/Explanation
\(\textbf{1(a)}\)
(i) 
Approximately symmetric about the \( x \)-axis graph of \( y^2 = x^3 \), including cusp/sharp point at \( (0, 0) \).
(ii) Approximately symmetric about the \( x \)-axis graph of \( y^2 = x^3 + 1 \) with approximately correct gradient at axes intercepts.
Some indication of the position of intersections at \( x = -1, y = \pm 1 \).
\(\textbf{1(b)}\)
(i) \((0, 1) \text{ and } (0, -1)\)
(ii) Any two from:
$y^2 = x^3$ has a cusp/sharp point, (the other does not)
graphs have different domains
$y^2 = x^3 + 1$ has points of inflection, (the other does not)
graphs have different $x$-axis intercepts (one goes through the origin, and the other does not)
graphs have different $y$-axis intercepts
\(\textbf{1(c)}\)
Any two form:
As $x \to \infty$, $y \to \pm \infty$
As $x \to \infty$, $y^2 = x^3 + b$ is approximated by $y^2 = x^3$ (or similar)
They have $x$-intercepts at $x = -\sqrt[3]{b}$
They have $y$-intercepts at $y = (\pm) \sqrt{b}$
They all have the same range
$y = 0$ (or $x$-axis) is a line of symmetry
They all have the same line of symmetry ($y = 0$)
They have one $x$-axis intercept
They have two $y$-axis intercepts
They have two points of inflexion
At $x$-axis intercepts, curve is vertical/infinite gradient
There is no cusp/sharp point at $x$-axis intercepts
\(\textbf{1(d)}\)
(i) Attempt to differentiate implicitly:
\(
2y \frac{dy}{dx} = 3x^2 + 1
\)
\(
\frac{dy}{dx} = \frac{3x^2 + 1}{2y} \quad \text{OR} \quad (\pm) 2\sqrt{x^3 + x} \frac{dy}{dx} = 3x^2 + 1
\)
\(
\frac{dy}{dx} = \pm \frac{3x^2 + 1}{2\sqrt{x^3 + x}}
\)
(ii) \(\text{Local minima/maxima occur when } \frac{dy}{dx} = 0
\)
\(
1 + 3x^2 = 0 \text{ has no (real) solutions (or equivalent).}
\)
so, no local minima/maxima exist.
\(\textbf{1(e)}\)
Attempts implicit differentiation on \(2y\frac{dy}{dx} = 3x^2 + 1\)
\(
2\left(\frac{dy}{dx}\right)^2 + 2y \frac{d^2y}{dx^2} = 6x
\)
\text{Recognizes that } \frac{d^2y}{dx^2} = 0
\(
\frac{dy}{dx} = \pm \sqrt{3x}
\)
\(
(\pm) \frac{3x^2 + 1}{2\sqrt{x^3 + x}} = (\pm) \sqrt{3x}
\)
\(
12x(x + x^3) = (1 + 3x^2)^2
\)
\(
12x^2 + 12x^4 = 9x^4 + 6x^2 + 1
\)
\(
3x^4 + 6x^2 – 1 = 0
\)
Attempt to use quadratic formula or equivalent:
\(
x^2 = \frac{-6 \pm \sqrt{48}}{6}
\)
\(
(x > 0 \implies) x = \frac{\sqrt{2\sqrt{3} – 3}}{3} \quad (p = 2, q = -3, r = 3)
\)
\(\textbf{1(f)}\)
(i) Attempt to find tangent line through (-1, -1)
\(
y + 1 = -\frac{3}{2}(x + 1) \quad \text{OR} \quad y = -1.5x – 2.5
\)
(ii) attempt to solve simultaneously with \(y^2 = x^3 + 2\)
obtain \(\left( \frac{17}{4}, -\frac{71}{8} \right)\)
\(\textbf{1(g)}\)
attempt to find equation of [QS]
\(
\frac{y – 1}{x + 1} = -\frac{79}{42} \; \left( = -1.88095\ldots \right)
\)
solve simultaneously with \(y^2 = x^3 + 2\)
\(
x = 0.28798\ldots \; \left( = \frac{127}{441} \right)
\)
\(
y = -1.4226\ldots \; \left( = \frac{13175}{9261} \right)
\)
(0.288, -1.42)