Hint

Minimise \(\displaystyle AP^{2}\) before taking a square root. Explain why the chosen stationary value gives the global minimum.

Worked solution

(a)

\(\displaystyle Q=(x-0)^{2}+(x^{2}-3)^{2}=x^{2}+x^{4}-6x^{2}+9=x^{4}-5x^{2}+9\).

Answer: \(Q=x^4-5x^2+9\).

(b)

\(\displaystyle Q'=4x^{3}-10x=2x(2x^{2}-5)\). For positive x it is negative before \(\displaystyle \sqrt{\frac{5}{2}}\) and positive afterwards. Hence Q decreases from the boundary \(\displaystyle x=0\), then increases after the unique interior minimum. The nearest point has \(\displaystyle x= \sqrt{\frac{5}{2}}= \frac{\sqrt{10}}{2}\) and \(\displaystyle y= \frac{5}{2}\). Minimising Q minimises AP since the square root is increasing.

Answer: \(\displaystyle P= (\frac{\sqrt{10}}{2}, \frac{5}{2})\).

(c)

At the point, \(\displaystyle Q= \frac{5}{2}+ (- \frac{1}{2})^{2}= \frac{11}{4}\), so \(\displaystyle AP= \frac{\sqrt{11}}{2}\). The line AP has slope \(\displaystyle \frac{\frac{5}{2}- 3}{\frac{\sqrt{10}}{2}}= - \frac{1}{\sqrt{10}}\). The parabola tangent slope is \(\displaystyle 2x=\sqrt{10}\); their product is −1, proving AP is normal.

Answer: Least distance \(\displaystyle \frac{\sqrt{11}}{2}\); AP slope \(\displaystyle - \frac{1}{\sqrt{10}}\), tangent slope \(\sqrt{10}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 2 marks

• 1 mark: Use both coordinate differences in the distance-squared formula.
• 1 mark: Expand to \(\displaystyle x^{4}-5x^{2}+9\).

Alternative methods: An equivalent factored distance expression with the correct expansion is accepted.

Common error: Minimising only the vertical separation.

(b) · 3 marks

• 1 mark: Differentiate and obtain the feasible positive stationary input \(\displaystyle \sqrt{\frac{5}{2}}\).
• 1 mark: Give the corresponding point \(\displaystyle (\frac{\sqrt{10}}{2}, \frac{5}{2})\).
• 1 mark: Justify a global minimum on \(\displaystyle x\ge 0\) and the use of squared distance.

Alternative methods: Use \(\displaystyle Q= (x^{2}- \frac{5}{2})^{2}+ \frac{11}{4}\) as a global bound, with its equality value feasible.

Common error: Selecting the stationary boundary \(\displaystyle x=0\) without comparing behaviour or ignoring \(\displaystyle x\ge 0\).

(c) · 3 marks

• 1 mark: Give the distance \(\displaystyle \frac{\sqrt{11}}{2}\) rather than its square.
• 1 mark: Find the joining-line and tangent slopes correctly.
• 1 mark: Use their product −1 to conclude perpendicularity.

Alternative methods: A zero dot product of direction vectors is accepted.

Common error: Reporting \(\displaystyle \frac{11}{4}\) as the distance or checking perpendicularity at a different point.

Total: 8 marks.