(i) Length of AB:
We have the equation \(4x^{1/2} = x + 3\). Let’s solve this equation:
\( 16x = (x + 3)^2 \)
\( 16x = x^2 + 6x + 9 \)
\( x^2 – 10x +9 = 0 \)
Now, we can solve this quadratic equation using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
For our equation, \(a = 1\), \(b = -10\), and \(c = 9\).
\( x = \frac{10 \pm \sqrt{(-10)^2 – 4(1)(9)}}{2(1)} \)
\( x = \frac{10 \pm \sqrt{100-36}}{2} \)
\( x = \frac{10 \pm \sqrt{64}}{2} \)
\( x = \frac{10 \pm 2\sqrt{64}}{2} \)
\( x = \frac{10\pm 8}{2} \)
\(\x=frac{10+8}{2}\)  and \(x= \frac{10-8}{2}\)
x=9 and x=1
So, we have two values for \(x\):9 and 1
Now, substitute these values into either the equation of the curve or the line to find the corresponding \(y\)-coordinates. Let’s use the equation of the curve:
\( y = 4x^{1/2} \)
For x=9
\( y = 4(9)^{1/2} \)
y=12
Similarly, for x=1
\( y = 4(1)^{1/2} \)
y=4
Now, we have the coordinates of points A and B. Use the distance formula to find the length of \(AB\):
\( \text{Length of } AB = \sqrt{(x_A – x_B)^2 + (y_A – y_B)^2} \)
\(AB=\sqrt{(9-1)^{2}+(12-4)^{2}}\)
\(AB=\sqrt{8^{2}+8^{2}}\)
\(AB=\sqrt{64+64}\)
\(AB=\sqrt{128}\)
\(AB=11.3\)
(ii) The tangent to the curve at point T is parallel to AB. For a parallel tangent, the slopes of the curve and the line are equal. Find the slope of the line \(y = x + 3\) (which is 1), and set it equal to the derivative of the curve’s equation with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(4x^{1/2}) \)
\( \frac{1}{2} \cdot 4x^{-1/2} = 1 \)
\( 2x^{-1/2} = 1 \)
\(x=4\)
Now substitute x into the curve equation:
\(y=4x^{\frac{1}{2}}\)
\(y=8\)
The coordinates \((4,8)\) are the coordinates of point T.
(iii)The equation of the normal is given by:
\( y – 8 = -1(x – 4) \)
Now, let’s eliminate \(y\) or \(x\):
\( -x + 12 = x + 3 \)
\( 2x = 9 \)
\( x = \frac{9}{2} \)
Now, substitute \(x\) back into the equation of the normal to find \(y\):
\( y – 8 = -1\left(\frac{9}{2} – 4\right) \)
\( y – 8 = -1\left(\frac{1}{2}\right) \)
\( y – 8 = -\frac{1}{2} \)
\( y = \frac{15}{2} \)
So, the point of intersection is \(\left(\frac{9}{2}, \frac{15}{2}\right)\), which is equivalent to \(\left(4^{\frac{1}{2}}, 7^{\frac{1}{2}}\right)\).