Quadratics: Question 3

Worked solution

Part (a): Forming the intersection equation

At any intersection point, both equations give the same $y$, so:

$x^2 - 5x + 11 = x + k$

Collect everything on one side:

$x^2 - 5x - x + 11 - k = 0$

$x^2 - 6x + (11 - k) = 0$

as required.

Part (b): Using the discriminant for a tangent

A line is a tangent to a curve when the quadratic formed by their intersection has a repeated root, i.e. discriminant $=0$.

Here $a=1$, $b=-6$, $c = 11-k$, so:

$b^2 - 4ac = 0$

$(-6)^2 - 4(1)(11-k) = 0$

$36 - 44 + 4k = 0$

$4k - 8 = 0$

$k = 2$

Part (c): Finding the point of tangency

Substitute $k=2$ back into the equation from part (a):

$x^2 - 6x + (11-2) = 0$

$x^2 - 6x + 9 = 0$

$(x - 3)^2 = 0$

$x = 3 \text{ (repeated root, confirming tangency)}$

Find $y$ using $L$: $y = x + k = 3 + 2 = 5$.

Check using $C$: $y = (3)^2 - 5(3) + 11 = 9 - 15 + 11 = 5$. ✓ Both equations agree.

So $L$ touches $C$ at $(3, 5)$.

Final answers

• (a) $x^2 - 6x + (11-k) = 0$ (shown)
• (b) $k = \boxed{2}$
• (c) Point of tangency $= \boxed{(3, 5)}$