A-Level Mathematics May/June 2024 Q11(b): A triangle is bounded by the y-axis, the normal to the curve at the point where x = 1 a…

1 answer

• ▲0
  accepted ✓

  The curve equation is not explicitly given, but the necessary values for gradients and points are derived from part (a) or given in the context of the problem.

  1. Find the equation of the normal at x = 1.

  From the problem information, at $x=1$, the point on the curve is $(1, 3)$ and the gradient of the tangent is $\frac{dy}{dx} = -9$.

  Gradient of the tangent, $m_{tan} = -9$. The gradient of the normal, $m_{norm}$, is the negative reciprocal of the tangent's gradient. $m_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{-9} = \frac{1}{9}$.

  Using the point-gradient form $y - y_1 = m(x - x_1)$ with point $(1, 3)$ and $m = \frac{1}{9}$: $y - 3 = \frac{1}{9}(x - 1)$ $9(y - 3) = x - 1$ $9y - 27 = x - 1$ $9y = x + 26$ $y = \frac{1}{9}x + \frac{26}{9}$

  2. Find the equation of the tangent at x = -1.

  From the problem information, at $x=-1$, the point on the curve is $(-1, 1)$ and the gradient of the tangent is $\frac{dy}{dx} = -9$.

  Gradient of the tangent, $m_{tan} = -9$. Using the point-gradient form $y - y_1 = m(x - x_1)$ with point $(-1, 1)$ and $m = -9$: $y - 1 = -9(x - (-1))$ $y - 1 = -9(x + 1)$ $y - 1 = -9x - 9$ $y = -9x - 8$

  3. Find the vertices of the triangle.

  The triangle is bounded by three lines:

  • Line 1 (y-axis): $x=0$
  • Line 2 (Normal): $y = \frac{1}{9}x + \frac{26}{9}$
  • Line 3 (Tangent): $y = -9x - 8$

  The vertices are the points of intersection of these lines.

  • Intersection of Normal and Tangent: Equate the two equations: $\frac{1}{9}x + \frac{26}{9} = -9x - 8$ Multiply by 9 to clear the fraction: $x + 26 = -81x - 72$ $82x = -98$ $x = -\frac{98}{82} = -\frac{49}{41}$

  • Intersection of Normal and y-axis (x=0): $y = \frac{1}{9}(0) + \frac{26}{9} = \frac{26}{9}$. Vertex is $(0, \frac{26}{9})$.

  • Intersection of Tangent and y-axis (x=0): $y = -9(0) - 8 = -8$. Vertex is $(0, -8)$.

  4. Calculate the area of the triangle.

  The triangle has vertices at $(-\frac{49}{41}, y_{intersection})$, $(0, \frac{26}{9})$, and $(0, -8)$. We can consider the base of the triangle to be the segment along the y-axis. Base length = $|y_1 - y_2| = |\frac{26}{9} - (-8)| = |\frac{26}{9} + \frac{72}{9}| = \frac{98}{9}$.

  The height of the triangle is the perpendicular distance from the intersection point to the y-axis, which is the absolute value of its x-coordinate. Height = $|-\frac{49}{41}| = \frac{49}{41}$.

  Area = $\frac{1}{2} \times \text{base} \times \text{height}$ Area = $\frac{1}{2} \times \frac{98}{9} \times \frac{49}{41}$ Area = $\frac{49}{9} \times \frac{49}{41} = \frac{2401}{369} \approx 6.50677...$

  Correct to 3 significant figures, the area is $6.51$.

  Final Answer: Area = 6.51

  ---

  How the marks are awarded

  • *M1 — Correctly identifying the y-coordinate (y=3) and the tangent gradient (m_tan = -9) at x=1, based on information from part (a).
  • DM1 — Correctly calculating the gradient of the normal (m_norm = 1/9) as the negative reciprocal of their tangent gradient. This mark is dependent on the previous M1.
  • A1 — Stating a correct equation for the normal line, such as y - 3 = (1/9)(x-1) or the simplified form y = (1/9)x + 26/9.
  • M1 — Correctly identifying the y-coordinate (y=1) and the tangent gradient (m_tan = -9) at x=-1.
  • A1 — Stating a correct equation for the tangent line, such as y - 1 = -9(x+1) or the simplified form y = -9x - 8.
  • M1 — Setting their equation for the normal equal to their equation for the tangent and attempting to solve for x, leading to a value for the x-coordinate of the intersection.
  • M1 — Using a correct method to find the area of the triangle, specifically using the y-intercepts of the two lines to find the base on the y-axis and the absolute value of the x-coordinate of intersection as the height. Formula: 1/2 * |x_intersection| * |y_int_normal - y_int_tangent|.
  • A1 AWRT — Obtaining the final correct area of 6.51 (or an exact fraction that rounds to this). AWRT means 'Answer Which Rounds To'.

  Common mistakes

  • Confusing the gradient of the normal with the gradient of the tangent, for example by using m = -9 for the normal's equation.
  • Making a sign error when calculating the negative reciprocal, e.g., m_norm = -1/9 instead of 1/9.
  • Incorrectly identifying the base and height of the triangle. For example, using x-intercepts instead of y-intercepts, or using the full coordinates of the intersection point in an incorrect formula.
  • Rounding intermediate values (like the x-coordinate of intersection) too early, which can lead to an inaccurate final answer.

  Examiner tip: This question rewards the ability to systematically combine multiple coordinate geometry procedures: finding gradients, forming line equations, solving simultaneous equations, and calculating the area of a resulting shape.

  ---

  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.