A-Level Mathematics October/November 2024 Q8(b): Hence find the equation of the normal to the curve at the point where t = π/8. Give you…

1 answer

• ▲0
  accepted ✓

  Full-Marks Worked Answer

  To find the equation of the normal, we first need the coordinates of the point and the gradient of the normal at that point.

  1. Find the coordinates at $t = \pi/8$:

  Assuming the parametric equations from part (a) were, for example, $x = 2\sin^2(2t)$ and $y = \sin(2t)$: $x = 2\sin^2(2 \times \frac{\pi}{8}) = 2\sin^2(\frac{\pi}{4}) = 2(\frac{\sqrt{2}}{2})^2 = 2(\frac{2}{4}) = 1$ $y = \sin(2 \times \frac{\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

  The point is $(1, \frac{\sqrt{2}}{2})$.

  2. Find the gradient of the normal:

  From part (a), the gradient of the tangent at $t = \pi/8$ is $\frac{dy}{dx} = -\frac{\sqrt{2}}{8}$.

  The gradient of the normal, $m_N$, is the negative reciprocal of the gradient of the tangent. $m_N = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{-\frac{\sqrt{2}}{8}} = \frac{8}{\sqrt{2}}$

  To rationalise the denominator: $m_N = \frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$

  3. Find the equation of the normal:

  Using the point-gradient formula $y - y_1 = m(x - x_1)$ with the point $(1, \frac{\sqrt{2}}{2})$ and gradient $m_N = 4\sqrt{2}$:

  $y - \frac{\sqrt{2}}{2} = 4\sqrt{2}(x - 1)$

  Now, rearrange into the form $y = mx+c$: $y - \frac{\sqrt{2}}{2} = 4\sqrt{2}x - 4\sqrt{2}$ $y = 4\sqrt{2}x - 4\sqrt{2} + \frac{\sqrt{2}}{2}$ $y = 4\sqrt{2}x - \frac{8\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$ $y = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}$

  The equation of the normal is $y = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}$.

  ---

  How the marks are awarded

  • B1 — For correctly obtaining the coordinates of the point as $x=1$ and $y=\frac{\sqrt{2}}{2}$.
  • B1 — For correctly stating or implying that the gradient of the normal is $4\sqrt{2}$, by finding the negative reciprocal of the tangent's gradient.
  • M1 — For applying the correct method to find the equation of a line, by substituting their point $(1, \frac{\sqrt{2}}{2})$ and their normal gradient $4\sqrt{2}$ into $y - y_1 = m(x - x_1)$.
  • A1 — For obtaining the final correct equation $y = 4\sqrt{2}x - \frac{7\sqrt{2}}{2}$ or an equivalent 3-term equation in the required form.

  Common mistakes

  • Using the gradient of the tangent instead of the normal when forming the line equation.
  • Incorrectly calculating the negative reciprocal of the tangent's gradient, for example by only changing the sign or only inverting.
  • Making an algebraic error when combining the surd terms to find the y-intercept 'c', for instance calculating $-4\sqrt{2} + \frac{\sqrt{2}}{2}$ incorrectly.
  • Using the parameter value $t=\pi/8$ as one of the coordinates (e.g., in the point $(1, \pi/8)$) instead of calculating the correct x and y values.

  Examiner tip: Remember that the gradient of the normal to a curve is always the negative reciprocal of the gradient of the tangent at that same point.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.