1. ## Difficult question

Sketch the curve $y^2=(x-a)(x-2a)^2$
Here I let a=1
Show that the normal to either branch of the curve at the point (2a,0) meets the curve again at the point where $x=a+\frac{1}{a}$.
From the geometry of the figure i know the statements are true but how do I prove it? Differentiating the equation here and plugging the values of x and y doesn't work here. Is there any other way to find the gradient of the normal, that is if i need to find it?
Thanks!

2. The curve is described by two functions:

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

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

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

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 $x \to 2a^+$ and $x \to 2a^-$which will be different due to the absolute value function in the denominator.

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

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

You can find the equation for the normal line using:

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

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

3. Originally Posted by calum
The curve is described by two functions:

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

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

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

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 $x \to 2a^+$ and $x \to 2a^-$which will be different due to the absolute value function in the denominator.

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

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

You can find the equation for the normal line using:

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

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

$
x = a + t^2,\; y = t\left(t^2-a\right)
$

so

$
\frac{dy}{dx} = \frac{3t^2-a}{2t}
$

the point of tangency is $x = 2a$ so $t = \sqrt{a}$ giving $\frac{dy}{dx} = \sqrt{a}$ as you said. The rest follows.

4. Danny, that's a really useful way of approaching the problem. None of my maths units have touched parametric equations in any depth . Are there any rules or tricks to identify the best way to parametrize the equation, because it just doesn't seem obvious to me from looking at the original equation?

5. Originally Posted by calum
Danny, that's a really useful way of approaching the problem. None of my maths units have touched parametric equations in any depth . Are there any rules or tricks to identify the best way to parametrize the equation, because it just doesn't seem obvious to me from looking at the original equation?
This is what I saw when I looked at the equation

$y^2 = (x-a)(x^2-2a)^2$ so $\left(\frac{y}{x-a}\right)^2 = x - a$ so if $x - a = t^2$ then $\left(\frac{y}{x-a}\right)^2 = t^2$ so $\frac{y}{x-a} = t$ (you can suppress the $\pm$ as $t$ can go negative) from which you can obtain the parameterization I obtained.