The curve C has equation $y = 4x + 3x^{rac{3}{2}} - 2x^2$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2

To verify that the point P(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:

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

Thus, we have:

$y = 16 + 24 - 32 = 8$

This confirms that the point P(4, 8) lies on C.

First, we find the slope of the normal line. We already calculated ( \frac{dy}{dx} ) at ( x = 4 ):

$\frac{dy}{dx} = 4 + \frac{9}{2}(4)^{\frac{1}{2}} - 4(4) = 4 + 18 - 16 = 6$

The slope of the normal is the negative reciprocal of the derivative:

$slope_{normal} = -\frac{1}{6}$.

Using the point-slope form for the point P(4, 8):

$y - 8 = -\frac{1}{6}(x - 4)$

Multiplying through by 6 to avoid fractions:

$6y - 48 = -x + 4$

Rearranging gives:

$x + 6y = 52$ ( \Rightarrow ) ( 3y = x + 20 ).

To find where the normal line intersects the x-axis, set ( y = 0 ) in the equation ( 3y = x + 20 ):

$3(0) = x + 20 \Rightarrow x = -20$.

Now we find the length PQ:

Using the distance formula: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the points P(4, 8) and Q(-20, 0):

$PQ = \sqrt{((-20) - 4)^2 + (0 - 8)^2}$ $= \sqrt{(-24)^2 + (-8)^2}$ = \sqrt{576 + 64} = \sqrt{640}\ As a simplified surd form, we have:

$$PQ = 8\sqrt{10}.$