A-Level Mathematics October/November 2024 Q9(a): The equation of a curve is y = ½k²x² - 2kx+2 and the equation of a line is y = kx+p, wh…

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  The equation of the curve is $y = \frac{1}{2}k^2x^2 - 2kx+2$. The equation of the line is $y = kx+p$. The point of intersection is $(\frac{5}{2}, \frac{1}{2})$.

  Since the point lies on the curve, we can substitute its coordinates into the curve's equation: $\frac{1}{2} = \frac{1}{2}k^2(\frac{5}{2})^2 - 2k(\frac{5}{2}) + 2$ $\frac{1}{2} = \frac{1}{2}k^2(\frac{25}{4}) - 5k + 2$ $\frac{1}{2} = \frac{25}{8}k^2 - 5k + 2$

  Multiply the entire equation by 8 to eliminate fractions: $4 = 25k^2 - 40k + 16$ $25k^2 - 40k + 12 = 0$

  We can solve this quadratic equation for $k$ by factorising: $(5k - 2)(5k - 6) = 0$ So, $k = \frac{2}{5}$ or $k = \frac{6}{5}$. Given the condition $0 < k < 1$, we must choose $k = \frac{2}{5}$.

  Now, since the point $(\frac{5}{2}, \frac{1}{2})$ also lies on the line, we substitute the coordinates and the value of $k$ into the line's equation to find $p$: $y = kx + p$ $\frac{1}{2} = (\frac{2}{5})(\frac{5}{2}) + p$ $\frac{1}{2} = 1 + p$ $p = \frac{1}{2} - 1 = -\frac{1}{2}$

  So, $k = \frac{2}{5}$ and $p = -\frac{1}{2}$.

  To find the other point of intersection, we equate the equations of the curve and the line with these values of $k$ and $p$. $\frac{1}{2}k^2x^2 - 2kx + 2 = kx + p$ $\frac{1}{2}(\frac{2}{5})^2x^2 - 2(\frac{2}{5})x + 2 = (\frac{2}{5})x - \frac{1}{2}$ $\frac{2}{25}x^2 - \frac{4}{5}x + 2 = \frac{2}{5}x - \frac{1}{2}$

  Rearrange to form a quadratic equation in $x$: $\frac{2}{25}x^2 - \frac{4}{5}x - \frac{2}{5}x + 2 + \frac{1}{2} = 0$ $\frac{2}{25}x^2 - \frac{6}{5}x + \frac{5}{2} = 0$

  Multiply by 50 to clear fractions: $4x^2 - 60x + 125 = 0$

  We know one solution is $x = \frac{5}{2}$, so $(2x-5)$ is a factor. $(2x-5)(2x-25) = 0$ The solutions are $x = \frac{5}{2}$ (the given point) and $x = \frac{25}{2}$.

  The other point of intersection has $x = \frac{25}{2}$. We find the corresponding $y$-coordinate using the simpler line equation: $y = \frac{2}{5}x - \frac{1}{2}$ $y = \frac{2}{5}(\frac{25}{2}) - \frac{1}{2}$ $y = 5 - \frac{1}{2} = \frac{9}{2}$

  Final Answer: The values are $k = \frac{2}{5}$ and $p = -\frac{1}{2}$. The other point of intersection is $(\frac{25}{2}, \frac{9}{2})$.

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  How the marks are awarded

  • M1* — Substituting the coordinates (5/2, 1/2) into the curve equation and simplifying to the three-term quadratic $25k^2 - 40k + 12 = 0$.
  • A1 — Solving the quadratic for k and selecting the correct value $k = 2/5$ based on the given condition $0 < k < 1$.
  • DM1* — Using the coordinates (5

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