Figure 6 shows a sketch of the curve C with parametric equations $x = 2 an t + 1$ $y = 2 ext{sec}^2 t + 3$ $-rac{ au}{4} \, ext{≤} \, t \, ext{≤} \, rac{ au}{3}$ The line l is the normal to C at the point P where $t = rac{ au}{4}$ - Edexcel - A-Level Maths Pure - Question 1 - 2021 - Paper 1

To show that all points satisfy $y = \frac{1}{2}(x - 1) + 5$:

1. From the parametric equations, start from:

   • $x = 2\tan(t) + 1$
   • Rearranging gives: $\tan(t) = \frac{x - 1}{2}$

2. Substitute this value into $y$:

   • $y = 2\sec^2(t) + 3$
   • Using the identity $\sec^2(t) = 1 + \tan^2(t)$, we have:

   $\sec^2(t) = 1 + \left(\frac{x - 1}{2}\right)^2 = 1 + \frac{(x - 1)^2}{4}$

3. Thus: $y = 2\left(1 + \frac{(x - 1)^2}{4}\right) + 3$

   Simplifying gives: $y = 2 + \frac{(x - 1)^2}{2} + 3 = \frac{1}{2}(x - 1)^2 + 5$

4. Therefore, all points on curve C satisfy the equation:
   $y = \frac{1}{2}(x - 1) + 5$

To find the values of $k$ such that the line intersects the curve at two distinct points:

1. Set the equations equal to each other:

   • $\frac{1}{2}x + k = 2\sec^2(t) + 3$
   • Substitute $x = 2\tan(t) + 1$:
     $\frac{1}{2}(2\tan(t) + 1) + k = 2\sec^2(t) + 3$

2. Rearranging gives: $\tan(t) + \frac{1}{2} + k = 2(1 + \tan^2(t)) + 3$

3. Transforming into a quadratic in terms of $\tan(t)$ leads to:

   • $2\tan^2(t) - \tan(t) + (2 + k - 3.5) = 0$

4. For this quadratic to have two distinct solutions, the discriminant must be greater than zero: $b^2 - 4ac > 0$

5. Use the discriminant:

   • $(-1)^2 - 4(2)(k - 1.5) > 0$

6. Simplifying leads to: $1 - 8(k - 1.5) > 0$ $1 - 8k + 12 > 0$ $13 - 8k > 0$

7. This results in: $k < \frac{13}{8}$

Therefore, the range of possible values for $k$ is: $k < 1.625$