June 2024 Paper 3 Q17
17 This question refers to the article on the Insert, “Tangents and normals to a quadratic curve”. The relevant extract (lines 25 to 28) is reproduced here.
For the curve \(y = x^2\), the coordinates of the point of intersection are not as simply related to the coordinates of A and B as in the case of the tangents. The equation of the normal at the point \((t, t^2)\) is \(y = -\tfrac{x}{2t} + t^2 + \tfrac{1}{2}\). The normals at points \((t_1, t_1^2)\) and \((t_2, t_2^2)\) cross when \(x = -2t_1t_2(t_1 + t_2)\) and \(y = t_1^2 + t_2^2 + t_1t_2 + \tfrac{1}{2}\).
Show that, for the curve \(y = x^2\), the equation of the normal at the point \((t, t^2)\) is \(y = -\tfrac{x}{2t} + t^2 + \tfrac{1}{2}\), as given in line 27. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) | M1 | 1.1 |
| Gradient of normal is \(-\dfrac{1}{2t}\) | M1 | 1.1 |
| \(y - t^2 = -\dfrac{1}{2t}(x - t)\) So \(y = -\tfrac{x}{2t} + t^2 + \tfrac{1}{2}\) | A1 | 2.1 |
| [3] |
Notes
M1: Accept \(2t\)
M1: For use of negative reciprocal oe
A1: For convincingly reaching given result
Any error seen is A0.
Alternative method for final mark
| Scheme | Marks | AO |
|---|---|---|
| For the given line when \(x = t\), \(y = -\tfrac{1}{2} + t^2 + \tfrac{1}{2} = t^2\) and the gradient is \(-\dfrac{1}{2t}\) | A1 |
Additional guidance
The first M1 is for differentiating to get either 2x or 2t.
The second M1 is for using the negative reciprocal to get -1/their gradient.
The A1 is for substituting into \((y - y_1) = m(x - x_1)\) or y= mx + c to get the required result or equivalent unsimplified. It must be convincing as the answer is given. Any error seen is A0.