Figure 6 shows a sketch of the curve C with parametric equations $x = 2 an t + 1$ $y = 2 ext{sec}^2 t + 3$ $-rac{ ext{π}}{4} ext{ } ext{≤ } t ext{ ≤ } rac{ ext{π}}{3}$ The line l is the normal to C at the point P where $t = rac{ ext{π}}{4}$. (a) Using parametric differentiation, show that an equation for l is $y = rac{1}{2} x + rac{17}{2}$ (b) Show that all points on C satisfy the equation $y = rac{1}{2} (x - 1)^2 + 5$ The straight line with equation $y = rac{1}{2} x + k$ where k is a constant intersects C at two distinct points. Find the range of possible values for k.

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

Question 2

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

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{ ext{π}}{4} ext{ } ext{≤ } t ext{ ≤ } rac{... show full transcript

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

Step 1

(a) Using parametric differentiation, show that an equation for l is

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Answer

To find the equation of the normal line l at point P where $t = \frac{\pi}{4}$ , we first need to compute the derivatives of the parametric equations.

Starting with the parametric equations:

$x = 2 \tan(t) + 1$
$y = 2 \sec^2(t) + 3$

We differentiate both x and y with respect to t:

The derivative of x with respect to t is: $\frac{dx}{dt} = 2 \sec^2(t)$ At $t = \frac{\pi}{4}$ , we find: $\frac{dx}{dt} = 2 \sec^2(\frac{\pi}{4}) = 2 \cdot 2 = 4$
The derivative of y with respect to t is: $\frac{dy}{dt} = 4 \sec^2(t) \tan(t)$ At $t = \frac{\pi}{4}$ , we obtain: $\frac{dy}{dt} = 4 \cdot 2 \cdot 1 = 8$

The slope of the tangent line at point P is: $m_{tangent} = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8}{4} = 2$

The slope of the normal line is the negative reciprocal: $m_{normal} = -\frac{1}{2}$

Now, we find the coordinates of point P when $t = \frac{\pi}{4}$ :

$x(P) = 2 \tan(\frac{\pi}{4}) + 1 = 3$
$y(P) = 2 \sec^2(\frac{\pi}{4}) + 3 = 7$

Thus, the normal line, $l$ , passing through point (3,7) with a slope of -1/2 can be written in point-slope form:

$y - y_1 = m_{normal} (x - x_1)$

$y - 7 = -\frac{1}{2}(x - 3)$

Rearranging gives: $y = -\frac{1}{2}x + \frac{3}{2} + 7$

Thus: $y = -\frac{1}{2}x + \frac{17}{2}$

This confirms the required equation for l.

Step 2

(b) Show that all points on C satisfy the equation

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Answer

To demonstrate that all points on curve C satisfy the equation

$y = \frac{1}{2} (x - 1)^2 + 5,$

we start by substituting the parametric expressions of x and y into the given equation:

Substitute $x = 2 \tan(t) + 1$ : $x - 1 = 2 \tan(t)$ Therefore: $(x - 1)^2 = (2 \tan(t))^2 = 4 \tan^2(t)$
Now substitute into y's parametric expression: $y = 2 \sec^2(t) + 3$

Thus, we can rewrite y using the identity $\sec^2(t) = 1 + \tan^2(t)$ : $y = 2(1 + \tan^2(t)) + 3 = 2 + 2 \tan^2(t) + 3 = 5 + 2 \tan^2(t)$

Now, we check if this matches: $y = 5 + \frac{1}{2}(4 \tan^2(t)) = 5 + 2 \tan^2(t)$

This indicates that: $y = \frac{1}{2} (x - 1)^2 + 5$

Thus proving that all points on C satisfy the equation.

Step 3

(c) Find the range of possible values for k.

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Answer

For the line given by the equation $y = \frac{1}{2}x + k$

to intersect curve C at two distinct points, the discriminant of the resulting quadratic equation must be positive.

Substitute $y$ from the line equation into the derived equation of curve C: $\frac{1}{2}(x - 1)^2 + 5 = \frac{1}{2}x + k$
Rearranging gives: $\frac{1}{2}(x^2 - 2x + 1) + 5 - \frac{1}{2}x - k = 0$

Simplifying further results in: $\frac{1}{2}x^2 - x + \left( rac{1}{2} + 5 - k \right) = 0$

or: $x^2 - 2x + (1 + 10 - 2k) = 0$

Denoting as: $a = 1 - 2k + 10$
The discriminant must be positive: $(-2)^2 - 4(1)(11 - 2k) > 0$

Therefore: $4 - 4(11 - 2k) > 0$

Simplifying leads to: $4k - 44 > 0$ which provides: $k > 11$
Additionally, the upward opening parabola must stay below the line with y-intercept k, ensuring: $k < 5$

Combining these inequalities gives: $11 < k < 5$

Thus, k must be in the range (11, 5), meaning there is no valid range of k; hence there would not be two intersection points.

(a) Using parametric differentiation, show that an equation for l is

(b) Show that all points on C satisfy the equation

(c) Find the range of possible values for k.

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