Consider the ellipse given by $x^2/4 + y^2 = 1$. Fix a point $z_0 = (x_0, y_0)$, where $x_0 > 0$ and $y_0 = \sqrt{1 - x_0 ^2/4}$. Given $z = (x, \sqrt{1 - x ^2/4})$ with $x > 0$, we put $$ \delta(z,z_0) = \left | \int_x ^ {x_0} \sqrt{1 - t ^2/4} \,dt \right |. $$
I'm trying to see if exists a constant $C> 0$ such that $\delta(z,z_0) \leq C |z-z_0|$, for all $z = (x, \sqrt{1 - x ^2/4})$ with $x > 0$.
My attempt:
$$\delta(z,z_0) = \left | \frac{1}{4} \left ( \sqrt{4 - x^2}\,x - \sqrt{4 - x_0 ^2}\, x_0 \right ) + \arcsin (x/2) - \arcsin(x_0/2) \right |$$ and $$ |z - z_0| = \left ( (x-x_0)^2 + \frac{1}{4} \left ( \sqrt{4 - x^2} - \sqrt{4 - x_0 ^2}\right )^2 \right )^{1/2}. $$
Is there any relation between the difference of $\arcsin$ that might help? Or another way to approach this problem?
Thank you.
This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question
$$ \delta(z,z_0) = \left | \int_x ^ {x_0} \sqrt{1 - t ^2/4} \,dt \right | \leq \left | \int_x ^ {x_0} 1 \,dt \right | \\ \leq |x-x_0|\leq |z-z_0|$$
$\Rightarrow C=1$ because $ \delta(z,z_0)\leq 1*|z-z_0|$
For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$\lim_{x\to 2} \frac{\delta((x,\sqrt{1-x^2/4}),(2,0))}{|(x,\sqrt{1-x^2/4})-(2,0)|}$$ Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ \delta(z,z_0)\geq C* |z-z_0|\forall z,z_0$