A-Level Mathematics October/November 2024 Q4: Show that the curve with equation x² – 3xy-40 = 0 and the line with equation 3x+y+k = 0…

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  To determine if the curve and the line meet, we solve their equations simultaneously.

  Curve: $x^2 - 3xy - 40 = 0$ --- (1) Line: $3x + y + k = 0$ --- (2)

  From the linear equation (2), we can express y in terms of x: $y = -3x - k$

  Now, substitute this expression for y into the equation of the curve (1): $x^2 - 3x(-3x - k) - 40 = 0$

  Expand the brackets and simplify the equation: $x^2 + 9x^2 + 3kx - 40 = 0$ $10x^2 + 3kx - 40 = 0$

  For the line and the curve to meet, this quadratic equation in x must have real roots. This occurs when the discriminant, $\Delta = b^2 - 4ac$, is greater than or equal to zero.

  For the quadratic $10x^2 + (3k)x - 40 = 0$, we have: $a = 10$, $b = 3k$, $c = -40$

  Now, we calculate the discriminant: $\Delta = (3k)^2 - 4(10)(-40)$ $\Delta = 9k^2 + 1600$

  To determine if the roots are always real, we must analyse the value of the discriminant for all values of k. For any real value of k, $k^2 \ge 0$. Therefore, $9k^2 \ge 0$. It follows that $9k^2 + 1600 \ge 1600$.

  Since 1600 > 0, the discriminant $\Delta = 9k^2 + 1600$ is always positive.

  Because the discriminant is always positive, the quadratic equation $10x^2 + 3kx - 40 = 0$ always has two distinct real roots for x. This means the line and the curve always intersect at two distinct points. Therefore, the curve and the line meet for all values of the constant k.

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  How the marks are awarded

  • *M1 — Awarded for correctly rearranging the linear equation to make y the subject ($y = -3x - k$) and substituting it into the curve's equation, $x^2 - 3x(-3x - k) - 40 = 0$.
  • A1 — Awarded for correctly expanding and simplifying the substituted equation to obtain the quadratic $10x^2 + 3kx - 40 = 0$.
  • DM1 — Awarded for attempting to find the discriminant ($b^2-4ac$) of the 3-term quadratic in x. This requires correctly identifying $a=10$, $b=3k$, and $c=-40$. This mark is dependent on the first M1 mark.
  • A1 — Awarded for correctly calculating the discriminant as $9k^2 + 1600$.
  • A1 FT — Awarded for stating that $9k^2 + 1600 > 0$ (or $\ge 0$) with a valid reason (e.g., $k^2 \ge 0$) and concluding that the line and curve therefore always meet.

  Common mistakes

  • Sign errors during substitution and expansion, for example writing $x^2 - 3x(-3x-k) = x^2 - 9x^2 - 3kx$, which leads to an incorrect quadratic.
  • Errors in calculating the discriminant, such as mishandling the negative sign in 'c': $(3k)^2 - 4(10)(40) = 9k^2 - 1600$. This leads to an incorrect conclusion that the line and curve do not always meet.
  • Providing an incomplete final argument. For instance, finding the discriminant is $9k^2 + 1600$ but failing to explain why this expression is always positive (i.e., because $k^2 \ge 0$).
  • Incorrectly identifying the coefficients for the discriminant, for example using $b=3$ instead of $b=3k$.

  Examiner tip: To determine if a line and a curve always intersect, combine their equations to form a single quadratic and then prove its discriminant ($b^2-4ac$) is always greater than or equal to zero.

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