proving or disproving that two tangent lines are parallel to a curve Im trying to prove or disprove that given the function, $f(x)=0.5\sqrt{1-x^{2}}$,
There are two different tangent lines to $f(x)$ so they are parallel.
I tried to derivative but with no success.
 A: $y=f(x)=0.5\sqrt{1-x^2}\implies 4y^2=1-x^2$ with $y\ge0$
Let us find the condition of $y=mx+c$ being a tangent to the curve
So, $4y^2=1-\left(\frac{y-c}m\right)^2\implies y^2(1-4m^2)+2cy+(m^2-c^2)=0$
This is a Quadratic Equation in $y$
For tangency, both the root must be same  i..e, the discriminant must be $0$ 
So, $$(2c)^2=4(1-4m^2)(m^2-c^2)$$
$$\implies c^2=m^2-4m^4+4m^2c^2+c^2\implies 4c^2=1-4m^2\text{ if } m \text{ is finite and non-zero}$$ 
$c=\pm\sqrt{1-4m^2}$
So, $y=\frac{2c}{1-4m^2}=\frac1{2c}>0\implies c\ge 0$ so $c\not<0$ 
We won't get parallel tangents for finite and non-zero $m$
If $m=0, y=c \implies c\ge 0$
We won't get parallel tangents here.
If $\frac1m=0, y=0\implies 1-x^2=0\implies x=\pm1$
Evidently, the tangents passing through $(\pm1,0)$  are the only parallel tangents 
A: I am not sure about the answer! Let's go through the details.
The derivative is $-\frac{0.5x}{\sqrt{1-x^2}}$. This is defined only when $|x|\lt 1$.
We we try to find $s$ and $t$ such that $-\frac{0.5s}{\sqrt{1-s^2}}=-\frac{0.5t}{\sqrt{1-t^2}}$, we quickly find that if $|s|\lt 1$ and $|t|\lt 1$, we must have $s=t$. So there cannot be parallel tangent lines for distinct $x$ in the interval $(-1,1)$. 
By manipulation of $y=0.5\sqrt{1-x^2}$ (square both sides), we find that our curve is the top half of a standard ellipse. At the points $x=1$ and $x=-1$, the full ellipse has parallel vertical tangent lines. Whether these are tangent lines to the half ellipse depends on what definition we use for tangent line. 
