Home / IB Mathematics SL 5.4 Tangents and normal AI HL Paper 1- Exam Style Questions

IB Mathematics SL 5.4 Tangents and normal AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI HL Papers 1 - All Topics

Question

An astrophysicist uses the quadratic function \(y=x^2\) to model the reflective surface of a parabolic telescope mirror.
(a) Determine the equation of the normal line to the mirror’s surface at the point where \(x=2\).
A beam of light originating from a celestial object at position \((0, 10)\) strikes the mirror at the point \((2, 4)\).
(b) Calculate the gradient of the path taken by this light beam.
(c) Find the magnitude of the angle between the light beam and the normal to the mirror at the point of impact.

Most-appropriate topic codes:

• SL 5.4: Tangents and normals at a given point — part (a)
• SL 2.1: Gradient of a straight line — part (b)
• AHL 3.13: Angle between two lines — part (c)
▶️ Answer/Explanation
Detailed solution

(a)
\(y = x^2 \Rightarrow \frac{dy}{dx} = 2x\).
At \(x=2\), gradient of tangent \(m_T = 2(2) = 4\).
Gradient of normal \(m_N = -\frac{1}{4}\).
Equation of normal using point \((2, 4)\):
\(y – 4 = -\frac{1}{4}(x – 2)\)
\(y = -\frac{1}{4}x + \frac{9}{2}\).

(b)
Ray connects \((0, 10)\) and \((2, 4)\).
Gradient \(m_R = \frac{4 – 10}{2 – 0} = \frac{-6}{2} = -3\).

(c)
Use the formula for the angle \(\theta\) between two lines with gradients \(m_1\) and \(m_2\):
\(\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right|\)
Here \(m_1 = -3\) (ray) and \(m_2 = -0.25\) (normal).
\(\tan \theta = \left| \frac{-3 – (-0.25)}{1 + (-3)(-0.25)} \right| = \left| \frac{-2.75}{1 + 0.75} \right| = \left| \frac{-2.75}{1.75} \right| = \frac{11}{7}\).
\(\theta = \arctan(\frac{11}{7}) \approx 1.00\) radian (or \(57.5^{\circ}\)).

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