A-Level Edexcel Maths Pure Graphs of Functions: 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$ The line l is the normal to C at the point P where $t = \frac{\pi}{4}$ (a) Using parametric differentiation, show that an equation for l is y = $\frac{1}{2}x + \frac{17}{2}$ (b) Show that all points on C satisfy the equation y = $\frac{1}{2}(x - 1)^2 + 5$ The straight line with equation y = $\frac{1}{2}x + k$ where k is a constant intersects C at two distinct points - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

Question 16

$Figure-6-shows-a-sketch-of-the-curve-C-with-parametric-equations--$x-=-2--an-t-+-1$--y-=-2--ext{sec}^2-t-+-3$--The-line-l-is-the-normal-to-C-at-the-point-P-where-$t-=-\frac{\pi}{4}$--(a)-Using-parametric-differentiation,-show-that-an-equation-for-l-is--y-=-$\frac{1}{2}x-+-\frac{17}{2}$--(b)-Show-that-all-points-on-C-satisfy-the-equation--y-=-$\frac{1}{2}(x---1)^2-+-5$--The-straight-line-with-equation--y-=-$\frac{1}{2}x-+-k$-where-k-is-a-constant--intersects-C-at-two-distinct-points-Edexcel-A-Level Maths Pure-Question 16-2022-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$ The line l is the normal to C at the point P where $t ... 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$ The line l is the normal to C at the point P where $t = \frac{\pi}{4}$ (a) Using parametric differentiation, show that an equation for l is y = $\frac{1}{2}x + \frac{17}{2}$ (b) Show that all points on C satisfy the equation y = $\frac{1}{2}(x - 1)^2 + 5$ The straight line with equation y = $\frac{1}{2}x + k$ where k is a constant intersects C at two distinct points - Edexcel - A-Level Maths Pure - Question 16 - 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 to the curve at the given point, we first need to calculate the derivative of the parametric equations. The parametric equations given are:

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

The derivatives are:

dy/dt = 4 \sec^2 t \tan t

Next, we will find the slope of the tangent line at the point where $t = \frac{\pi}{4}$ :

At $t = \frac{\pi}{4}$ :

$dx/dt \bigg|_{t=\frac{\pi}{4}} = 2 \sec^2(\frac{\pi}{4}) = 4,$

Thus, the slope of the tangent line, $m_t$ , is:

The slope of the normal line, $m_n$ , is the negative reciprocal of the tangent slope:

To find the coordinates of point P, we substitute $t = \frac{\pi}{4}$ into the parametric equations:

$x = 2 \tan(\frac{\pi}{4}) + 1 = 3,$

The normal line equation in point-slope form is:

$y - y_0 = m_n(x - x_0)$

Substituting in the coordinates and the slope:

$y - 5 = -\frac{1}{2}(x - 3)$

Rearranging gives:

$y = -\frac{1}{2}x + \frac{3}{2} + 5 = -\frac{1}{2}x + \frac{13}{2},$

which can also be expressed as:

$y = \frac{1}{2}x + \frac{17}{2}$ .

Step 2

Show that all points on C satisfy the equation

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Answer

To show that all points on the curve C satisfy the equation

y = $\frac{1}{2}(x - 1)^2 + 5$ , we will first express y in terms of x using the original parametric equations:

From the x equation: $x = 2 \tan t + 1$ Solving for $an t$ gives: $\tan t = \frac{x - 1}{2}$
Now, we will substitute $an t$ into the y equation: $y = 2 \sec^2 t + 3$$$$\sec^2 t = 1 + \tan^2 t = 1 + \left(\frac{x - 1}{2}\right)^2 = 1 + \frac{(x - 1)^2}{4}$ Thus, $y = 2\left(1 + \frac{(x - 1)^2}{4}\right) + 3$
Simplifying: $y = 2 + \frac{(x - 1)^2}{2} + 3 = \frac{1}{2}(x - 1)^2 + 5$

This confirms that all points on curve C satisfy the equation.

Step 3

Find the range of possible values for k

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Answer

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

Substitute $y = \frac{1}{2}x + k$ into the equation of curve C: $\frac{1}{2}(x - 1)^2 + 5 = \frac{1}{2}x + k$
Rearranging gives: $\frac{1}{2}(x - 1)^2 - \frac{1}{2}x + (5 - k) = 0$
Multiply through by 2 to eliminate the fraction: $(x - 1)^2 - x + 2(5 - k) = 0$
The discriminant of this quadratic must be positive for there to be two distinct solutions: [\Delta = b^2 - 4ac = (-1)^2 - 4(1)(2(5 - k)) > 0] Which simplifies to: $1 - 8(5 - k) > 0$
Solving: $1 - 40 + 8k > 0$ $8k > 39$ $k > \frac{39}{8} = 4.875$
Thus, for the range of k ensuring two distinct intersection points: $k < 5$ So, The final result is: $4.875 < k < 5$

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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