The curve C has equation $y = 4x + 3x^{rac{3}{2}} - 2x^2, \, x > 0.$ (a) Find an expression for $rac{dy}{dx}.$ (b) Show that the point $P(4, 8)$ lies on $C.$ (c) Show that an equation of the normal to $C$ at the point $P$ is $3y = x + 20.$ The normal to $C$ at $P$ cuts the x-axis at the point $Q.$ (d) Find the length $PQ,$ giving your answer in a simplified surd form. - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2

To verify if the point $(4, 8)$ lies on the curve, we substitute $x = 4$ into the equation:

$y = 4(4) + 3(4)^{\frac{3}{2}} - 2(4^2)$

Calculating each term gives:

1. $4(4) = 16$.
2. $3(4)^{\frac{3}{2}} = 3(8) = 24$.
3. $-2(4^2) = -32$.

So we get:

$y = 16 + 24 - 32 = 8$

Thus, the point $(4, 8)$ does indeed lie on $C.$

First, we calculate the slope of the tangent line at $x=4$:

Using our prior expression for ( \frac{dy}{dx} ):

$\frac{dy}{dx} = 4 + \frac{9}{2}(4)^{\frac{1}{2}} - 4(4) = 4 + \frac{9}{2}(2) - 16$

This simplifies to:

$\frac{dy}{dx} = 4 + 9 - 16 = -3$

The slope of the normal is the negative reciprocal:

$\text{slope of normal} = \frac{1}{3}$

Using the point-slope form of the line:

$y - 8 = \frac{1}{3}(x - 4)$

Rearranging gives:

$3y - 24 = x - 4 \Rightarrow 3y = x + 20$

To find the coordinates of point $Q$, we set $y=0$ in the equation of the normal:

$0 = \frac{1}{3}x + \frac{64}{3}$

Solving for $x$:

$x = -20$

Now, we find $PQ$:

Using the distance formula:

$PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} = \sqrt{(-20 - 4)^2 + (0 - 8)^2}$

Calculating gives:

$PQ = \sqrt{(-24)^2 + (-8)^2} = \sqrt{576 + 64} = \sqrt{640} = 8\sqrt{10}$.