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**calum** The curve is described by two functions:

$\displaystyle y = \pm \sqrt{(x-a)(x-2a)^2}$

Find the derivative, which will be, for the positive square root function:

$\displaystyle y' = \frac{(x-2a)^2 + 2(x-a)(x-2a)}{\sqrt{x-a} |x-2a|}$

y' is not defined at x = 2a because the function is not differentiable at this point (although it is continuous). Nevertheless you can find the limits of y' as $\displaystyle x \to 2a^+$ and $\displaystyle x \to 2a^-$which will be different due to the absolute value function in the denominator.

$\displaystyle \lim_{x \to 2a^+} \frac{(x-2a)^2 + 2(x-a)(x-2a)}{\sqrt{x-a} |x-2a|} = \lim_{x \to 2a^+} \sqrt{x-a} + \frac{x-2a}{2 \sqrt{x-a}} = \sqrt{a}$

Of course, this is the gradient of the tangent, so the gradient of the normal will be $\displaystyle \frac{-1}{\sqrt{a}} $

You can find the equation for the normal line using:

$\displaystyle (y-0) = \frac{-1}{\sqrt{a}} (x - 2a)$

Now it should be obvious how to show that the normal line intersects the positive square root function at the given point.