Find the point on ellipse $4x^2+y^2=4$ that is furthest from $(1,0)$ The equation of an ellipse is $4x^2 + y^2 = 4$. Find the furthest point $(x,y)$ from $(1,0)$. 

Additional information from comment: First, I transformed this equation into the general formula and the result I got was: $$\frac {x^2}1 + \frac{y^2}{2^2} = 1.$$ 

My professor used this formula to find the distance between the 2 points: $$\sqrt{(x-1)^2 + (y-0)^2}$$ but I don't understand where this formula comes from.
The diagram below depicts the ellipse, and the right triangle involved in the distance formula:



In this diagram, the point on the ellipse has coordinates $(x, y)$ in general.  (Here, I've chosen $x = -\frac12$ and $y = \sqrt{3}$, but it could be any point on the ellipse.)  The two legs are represented by the blue line segment, which has length $|x-1|$, and the green line segment, which has length $|y|$.  The hypotenuse—the orange line segment—is the desired distance $r$.

From Pythagoras, we have 

$$
r^2 = (|x-1|)^2+(|y|)^2 = (x-1)^2+y^2
$$

where we can get rid of the absolute value because any real value $z$ and its absolute value $|z|$ have the same square.  Take the square root of both sides (since $r \geq 0$), and we get

$$
r = \sqrt{(x-1)^2+y^2}
$$
Let $d$ be the distance from the point $(1,0)$ to a point on the ellipse $(x,y)$, then $d^2 = (x-1)^2+y^2= (x-1)^2 + 4-4x^2= x^2-2x+1+4-4x^2 = -3x^2-2x+5 $. This is a quadratic function and you know it attains a max at $x = -\frac{b}{2a} = - \frac{-2}{2(-3)} = -\frac{1}{3}$. Thus $d^2_{\text{max}} = -3\left(\frac{-1}{3}\right)^2-2\left(\frac{-1}{3}\right)+5 = \frac{16}{3}$ . So $d_{\text{max}} = \dfrac{4}{\sqrt{3}}$. Hope this helps !