OCR MEI Paper 3 2024 June — Question 17 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyEasy -1.2 This is a straightforward application of finding a normal line equation using parametric point notation. Students need to differentiate y=x², find the gradient at (t,t²), take the negative reciprocal, and substitute into point-slope form. It's routine calculus with algebraic manipulation but requires no problem-solving insight—essentially a 'show that' verification exercise that's easier than average.
Spec1.07m Tangents and normals: gradient and equations

17 Show that, for the curve \(y = x ^ { 2 }\), the equation of the normal at the point \(\left( t , t ^ { 2 } \right)\) is \(y = - \frac { x } { 2 t } + t ^ { 2 } + \frac { 1 } { 2 }\), as given in line 27.

Question 17:
AnswerMarks Guidance
Working/AnswerMarks Guidance
Differentiate to get either \(2x\) or \(2t\)M1
Use negative reciprocal to get \(-\frac{1}{\text{their gradient}}\)M1
Substitute into \((y - y_1) = m(x - x_1)\) or \(y = mx + c\) to get required result or equivalent unsimplifiedA1 Answer is given — must be convincing; any error scores A0
## Question 17:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Differentiate to get either $2x$ or $2t$ | M1 | |
| Use negative reciprocal to get $-\frac{1}{\text{their gradient}}$ | M1 | |
| Substitute into $(y - y_1) = m(x - x_1)$ or $y = mx + c$ to get required result or equivalent unsimplified | A1 | Answer is given — must be convincing; any error scores A0 |

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17 Show that, for the curve $y = x ^ { 2 }$, the equation of the normal at the point $\left( t , t ^ { 2 } \right)$ is $y = - \frac { x } { 2 t } + t ^ { 2 } + \frac { 1 } { 2 }$, as given in line 27.

\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q17 [3]}}