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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