A-Level Edexcel Maths Pure Data Presentation: Figure 6 shows a sketch of the c

Figure 6 shows a sketch of the curve C with parametric equations $x = 2 an t + 1$ y = 2 ext{sec}^2 t + 3$ $-rac{rac{eta}{4}} ext{ } ext{ } rac{eta}{3} ext{ } | ext{}| | ext{}| ext{ }$ (a) The line l is the normal to C at the point P where $t = rac{eta}{4}$ (b) Show that all points on C satisfy the equation y = rac{1}{2} (x - 1)^2 + 5 (5) The straight line with equation y = rac{1}{2} (x - 1) + k ext{ where k is a constant} intersects C at two distinct points - Edexcel - A-Level Maths Pure - Question 1 - 2022 - Paper 2

Question 1

$Figure-6-shows-a-sketch-of-the-curve-C-with-parametric-equations--$x-=-2--an-t-+-1$--y-=-2--ext{sec}^2-t-+-3$--$--rac{-rac{eta}{4}}--ext{-}--ext{-}--rac{eta}{3}--ext{-}-|-ext{}|-|-ext{}|--ext{-}$-(a)-The-line-l-is-the-normal-to-C-at-the-point-P-where-$t-=--rac{eta}{4}$--(b)-Show-that-all-points-on-C-satisfy-the-equation--y-=--rac{1}{2}-(x---1)^2-+-5--(5)--The-straight-line-with-equation--y-=--rac{1}{2}-(x---1)-+-k--ext{-where-k-is-a-constant}--intersects-C-at-two-distinct-points-Edexcel-A-Level Maths Pure-Question 1-2022-Paper 2.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{rac{eta}{4}} ext{ } ext{ } rac{eta}{3} e... 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$ $-rac{rac{eta}{4}} ext{ } ext{ } rac{eta}{3} ext{ } | ext{}| | ext{}| ext{ }$ (a) The line l is the normal to C at the point P where $t = rac{eta}{4}$ (b) Show that all points on C satisfy the equation y = rac{1}{2} (x - 1)^2 + 5 (5) The straight line with equation y = rac{1}{2} (x - 1) + k ext{ where k is a constant} intersects C at two distinct points - Edexcel - A-Level Maths Pure - Question 1 - 2022 - Paper 2

Step 1

Using parametric differentiation, show that an equation for l is

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Answer

To find the normal line l, we first need to find dy/dx for the curve C. We have:

From the parametric equations, we differentiate:
- $x = 2 an t + 1$ gives us $rac{dx}{dt} = 2 ext{sec}^2 t$
- $y = 2 ext{sec}^2 t + 3$ gives us $rac{dy}{dt} = 4 ext{sec}^2 t an t$ .
Thus, the slope of the curve at point P ( $t = rac{eta}{4}$ ) is: $rac{dy}{dx} = rac{ rac{dy}{dt}}{ rac{dx}{dt}} = rac{4 ext{sec}^2( rac{eta}{4}) an( rac{eta}{4})}{2 ext{sec}^2( rac{eta}{4})} = 2 an( rac{eta}{4}).$
The slope of the normal line l, being the negative reciprocal, is: $m_l = - rac{1}{2 an( rac{eta}{4})}.$
The coordinates of point P when $t = rac{eta}{4}$ are found from the parametric equations:
- $x = 2 an( rac{eta}{4}) + 1$
- $y = 2 ext{sec}^2( rac{eta}{4}) + 3$ .
Therefore, the equation for line l using point-slope form ( $y - y_1 = m(x - x_1)$ ) is: $y - (2 ext{sec}^2( rac{eta}{4}) + 3) = - rac{1}{2 an( rac{eta}{4})}(x - (2 an( rac{eta}{4}) + 1)).$

Step 2

Show that all points on C satisfy the equation

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Answer

To find the Cartesian equation relating x and y, we can express t in terms of x:

From the equation $x = 2 an(t) + 1$ , we get: $an(t) = rac{x - 1}{2}.$
Substituting into the y equation: $y = 2 ext{sec}^2(t) + 3 = 2igg(1 + an^2(t)igg) + 3 = 2igg(1 + ig( rac{x - 1}{2}ig)^2igg) + 3.$
Simplifying: $y = 2igg(1 + rac{(x - 1)^2}{4}igg) + 3 = rac{1}{2}(x - 1)^2 + 5.$
This proves that all points (x, y) on C satisfy the equation: $y = rac{1}{2} (x-1)^2 + 5.$

Step 3

Find the range of possible values for k.

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Answer

To find the range of k for which the line intersects C at two distinct points:

Set the expressions for y equal: $rac{1}{2}(x - 1)^2 + 5 = rac{1}{2}(x - 1) + k.$
Rearranging gives: $rac{1}{2}(x - 1)^2 - rac{1}{2}(x - 1) + (5 - k) = 0.$
This is a quadratic equation in $x$ , and for it to have two distinct roots, the discriminant must be positive: $b^2 - 4ac > 0$ Here, $a = rac{1}{2}, b = - rac{1}{2}, c = 5 - k.$
The discriminant is:

ightarrow rac{1}{4} - 2(5 - k) > 0 ightarrow rac{1}{4} < 10 - 2k$$

Simplifying:

ightarrow 2k < rac{40 - 1}{4} ightarrow 8k < 39 ightarrow k < rac{39}{8} = 4.875.$$

For the other bound, we also have:

ightarrow k > 0.125.$$

Thus, the range of k is: $0.125 < k < 4.875.$

Using parametric differentiation, show that an equation for l is

Show that all points on C satisfy the equation

Find the range of possible values for k.

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